Dans toute la page, même si ce n'est pas indiqué, les variables prennent des valeurs telles que les calculs soient définis.
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Expressions irrationnellesExpressions algébriques/Exercices/Expressions irrationnelles », n'a pu être restituée correctement ci-dessus.
Rendre rationnel le dénominateur des expressions suivantes :
a)
1
2
2
+
3
+
10
+
15
{\displaystyle {\frac {1}{2{\sqrt {2}}+{\sqrt {3}}+{\sqrt {10}}+{\sqrt {15}}}}}
b)
1
1
+
2
3
+
4
3
{\displaystyle {\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}}
c)
1
2
−
3
+
5
−
7
{\displaystyle {\frac {1}{{\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}}-{\sqrt {7}}}}}
d)
1
2
3
+
3
3
+
5
3
{\displaystyle {\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}}
a) Corrigé.
1
2
2
+
3
+
10
+
15
=
(
2
2
+
10
)
−
(
3
+
15
)
[
(
2
2
+
10
)
+
(
3
+
15
)
]
[
(
2
2
+
10
)
−
(
3
+
15
)
]
=
2
2
+
10
−
3
−
15
(
2
2
+
10
)
2
−
(
3
+
15
)
2
=
2
2
+
10
−
3
−
15
(
8
+
8
5
+
10
)
−
(
3
+
6
5
+
15
)
=
2
2
+
10
−
3
−
15
2
5
=
(
2
2
+
10
−
3
−
15
)
5
2
5
5
=
2
10
+
5
2
−
15
−
5
3
10
{\displaystyle {\begin{aligned}{\frac {1}{2{\sqrt {2}}+{\sqrt {3}}+{\sqrt {10}}+{\sqrt {15}}}}&={\frac {(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})}{\left[(2{\sqrt {2}}+{\sqrt {10}})+({\sqrt {3}}+{\sqrt {15}})\right]\left[(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})\right]}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(2{\sqrt {2}}+{\sqrt {10}})^{2}-({\sqrt {3}}+{\sqrt {15}})^{2}}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(8+8{\sqrt {5}}+10)-(3+6{\sqrt {5}}+15)}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{2{\sqrt {5}}}}\\&={\frac {(2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}){\sqrt {5}}}{2{\sqrt {5}}{\sqrt {5}}}}\\&={\frac {2{\sqrt {10}}+5{\sqrt {2}}-{\sqrt {15}}-5{\sqrt {3}}}{10}}\\\end{aligned}}}
b) Corrigé.
Dans l'identité remarquable :
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
)
=
a
3
+
b
3
+
c
3
−
3
a
b
c
{\displaystyle (a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)=a^{3}+b^{3}+c^{3}-3abc}
nous poserons
a
=
1
b
=
2
3
c
=
4
3
{\displaystyle a=1\qquad \qquad b={\sqrt[{3}]{2}}\qquad \qquad c={\sqrt[{3}]{4}}}
et multiplions le numérateur et le dénominateur par
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
{\displaystyle a^{2}+b^{2}+c^{2}-ab-ac-bc}
1
1
+
2
3
+
4
3
=
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
(
1
+
2
3
+
4
3
)
(
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
)
=
1
+
4
3
+
16
3
−
2
3
−
4
3
−
8
3
1
+
2
+
4
−
3
1
×
2
×
4
3
=
1
+
4
3
+
2
2
3
−
2
3
−
4
3
−
2
7
−
3
8
3
=
2
3
−
1
7
−
3
×
2
=
2
3
−
1
{\displaystyle {\begin{aligned}{\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left(1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {1+{\sqrt[{3}]{4}}+{\sqrt[{3}]{16}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-{\sqrt[{3}]{8}}}{1+2+4-3{\sqrt[{3}]{1\times 2\times 4}}}}\\&={\frac {1+{\sqrt[{3}]{4}}+2{\sqrt[{3}]{2}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-2}{7-3{\sqrt[{3}]{8}}}}\\&={\frac {{\sqrt[{3}]{2}}-1}{7-3\times 2}}\\&={\sqrt[{3}]{2}}-1\end{aligned}}}
c) Corrigé.
1
2
−
3
+
5
−
7
=
(
2
−
3
)
−
(
5
−
7
)
[
(
2
−
3
)
+
(
5
−
7
)
]
[
(
2
−
3
)
−
(
5
−
7
)
]
=
(
2
−
3
)
−
(
5
−
7
)
(
2
−
3
)
2
−
(
5
−
7
)
2
=
2
−
3
−
5
+
7
(
5
−
2
6
)
−
(
12
−
2
35
)
=
2
−
3
−
5
+
7
2
35
−
7
−
2
6
=
[
2
−
3
−
5
+
7
]
[
2
35
+
7
+
2
6
]
[
2
35
−
(
7
+
2
6
)
]
[
2
35
+
(
7
+
2
6
)
]
=
2
70
+
7
2
+
4
3
−
2
105
−
7
3
−
6
2
−
10
7
−
7
5
−
2
30
+
14
5
+
7
7
+
2
42
140
−
(
7
+
2
6
)
2
=
2
−
3
3
+
7
5
−
3
7
−
2
30
+
2
42
+
2
70
−
2
105
67
−
28
6
=
(
2
−
3
3
+
7
5
−
3
7
−
2
30
+
2
42
+
2
70
−
2
105
)
(
67
+
28
6
)
(
67
−
28
6
)
(
67
+
28
6
)
=
−
185
2
−
145
3
+
133
5
+
135
7
+
62
30
+
50
42
−
34
70
−
22
105
−
215
=
185
2
+
145
3
−
133
5
−
135
7
−
62
30
−
50
42
+
34
70
+
22
105
215
{\displaystyle {\begin{aligned}{\frac {1}{{\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}}-{\sqrt {7}}}}&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{\left[({\sqrt {2}}-{\sqrt {3}})+({\sqrt {5}}-{\sqrt {7}})\right]\left[({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})\right]}}\\&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{({\sqrt {2}}-{\sqrt {3}})^{2}-({\sqrt {5}}-{\sqrt {7}})^{2}}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{(5-2{\sqrt {6}})-(12-2{\sqrt {35}})}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{2{\sqrt {35}}-7-2{\sqrt {6}}}}\\&={\frac {\left[{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}\right]\left[2{\sqrt {35}}+7+2{\sqrt {6}}\right]}{\left[2{\sqrt {35}}-(7+2{\sqrt {6}})\right]\left[2{\sqrt {35}}+(7+2{\sqrt {6}})\right]}}\\&={\frac {2{\sqrt {70}}+7{\sqrt {2}}+4{\sqrt {3}}-2{\sqrt {105}}-7{\sqrt {3}}-6{\sqrt {2}}-10{\sqrt {7}}-7{\sqrt {5}}-2{\sqrt {30}}+14{\sqrt {5}}+7{\sqrt {7}}+2{\sqrt {42}}}{140-(7+2{\sqrt {6}})^{2}}}\\&={\frac {{\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}}{67-28{\sqrt {6}}}}\\&={\frac {\left({\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}\right)\left(67+28{\sqrt {6}}\right)}{\left(67-28{\sqrt {6}}\right)\left(67+28{\sqrt {6}}\right)}}\\&={\frac {-185{\sqrt {2}}-145{\sqrt {3}}+133{\sqrt {5}}+135{\sqrt {7}}+62{\sqrt {30}}+50{\sqrt {42}}-34{\sqrt {70}}-22{\sqrt {105}}}{-215}}\\&={\frac {185{\sqrt {2}}+145{\sqrt {3}}-133{\sqrt {5}}-135{\sqrt {7}}-62{\sqrt {30}}-50{\sqrt {42}}+34{\sqrt {70}}+22{\sqrt {105}}}{215}}\\\end{aligned}}}
d) Corrigé.
Dans l'identité remarquable :
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
)
=
a
3
+
b
3
+
c
3
−
3
a
b
c
{\displaystyle (a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)=a^{3}+b^{3}+c^{3}-3abc}
nous poserons
a
=
2
3
b
=
3
3
c
=
5
3
{\displaystyle a={\sqrt[{3}]{2}}\qquad \qquad b={\sqrt[{3}]{3}}\qquad \qquad c={\sqrt[{3}]{5}}}
et multiplions le numérateur et le dénominateur par
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
{\displaystyle a^{2}+b^{2}+c^{2}-ab-ac-bc}
1
2
3
+
3
3
+
5
3
=
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
(
2
3
+
3
3
+
5
3
)
(
a
2
+
b
2
+
c
2
−
a
b
−
a
c
−
b
c
)
=
4
3
+
9
3
+
25
3
−
6
3
−
10
3
−
15
3
2
+
3
+
5
−
3
2
×
3
×
5
3
=
4
3
+
9
3
+
25
3
−
6
3
−
10
3
−
15
3
10
−
3
30
3
{\displaystyle {\begin{aligned}{\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left({\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{2+3+5-3{\sqrt[{3}]{2\times 3\times 5}}}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{10-3{\sqrt[{3}]{30}}}}\end{aligned}}}
Nous allons maintenant utiliser l'identité remarquable :
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
{\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})}
en posant :
a
=
10
b
=
3
30
3
{\displaystyle a=10\qquad \qquad b=3{\sqrt[{3}]{30}}}
Nous continuerons ainsi :
=
(
4
3
+
9
3
+
25
3
−
6
3
−
10
3
−
15
3
)
(
a
2
+
a
b
+
b
2
)
(
10
−
3
30
3
)
(
a
2
+
a
b
+
b
2
)
=
(
4
3
+
9
3
+
25
3
−
6
3
−
10
3
−
15
3
)
(
100
+
30
3
3
0
+
9
900
3
)
)
10
3
−
(
3
30
3
)
3
=
100
4
3
+
60
15
3
+
18
450
3
+
100
9
3
+
90
10
3
+
27
300
3
+
100
25
3
+
150
6
3
+
45
180
3
−
100
6
3
−
30
180
3
−
54
25
3
−
100
10
3
−
30
300
3
−
90
9
3
−
100
15
3
−
30
450
3
−
135
4
3
1000
−
810
=
−
35
4
3
+
50
6
3
+
10
9
3
−
10
10
3
−
40
15
3
+
46
25
3
+
15
180
3
−
3
300
3
−
12
450
3
190
{\displaystyle {\begin{aligned}\qquad &={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)(a^{2}+ab+b^{2})}{\left(10-3{\sqrt[{3}]{30}}\right)(a^{2}+ab+b^{2})}}\\&={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)\left(100+30{\sqrt[{3}]{3}}0+9{\sqrt[{3}]{900}}\right))}{10^{3}-\left(3{\sqrt[{3}]{30}}\right)^{3}}}\\&={\frac {100{\sqrt[{3}]{4}}+60{\sqrt[{3}]{15}}+18{\sqrt[{3}]{450}}+100{\sqrt[{3}]{9}}+90{\sqrt[{3}]{10}}+27{\sqrt[{3}]{300}}+100{\sqrt[{3}]{25}}+150{\sqrt[{3}]{6}}+45{\sqrt[{3}]{180}}-100{\sqrt[{3}]{6}}-30{\sqrt[{3}]{180}}-54{\sqrt[{3}]{25}}-100{\sqrt[{3}]{10}}-30{\sqrt[{3}]{300}}-90{\sqrt[{3}]{9}}-100{\sqrt[{3}]{15}}-30{\sqrt[{3}]{450}}-135{\sqrt[{3}]{4}}}{1000-810}}\\&={\frac {-35{\sqrt[{3}]{4}}+50{\sqrt[{3}]{6}}+10{\sqrt[{3}]{9}}-10{\sqrt[{3}]{10}}-40{\sqrt[{3}]{15}}+46{\sqrt[{3}]{25}}+15{\sqrt[{3}]{180}}-3{\sqrt[{3}]{300}}-12{\sqrt[{3}]{450}}}{190}}\end{aligned}}}
Établir les relations suivantes :
a)
(
11
5
−
3
)
2
−
(
5
−
2
5
2
−
5
)
2
=
91
4
+
10
3
{\displaystyle \left({\frac {11}{5-{\sqrt {3}}}}\right)^{2}-\left({\frac {5-2{\sqrt {5}}}{2-{\sqrt {5}}}}\right)^{2}={\sqrt {{\frac {91}{4}}+10{\sqrt {3}}}}}
b)
2
9
+
65
19
−
3
=
19
+
3
2
9
−
65
{\displaystyle {\frac {2{\sqrt {9+{\sqrt {65}}}}}{{\sqrt {19}}-{\sqrt {3}}}}={\frac {{\sqrt {19}}+{\sqrt {3}}}{2{\sqrt {9-{\sqrt {65}}}}}}}
c)
8
+
2
10
+
2
5
+
8
−
2
10
+
2
5
=
2
(
5
+
1
)
{\displaystyle {\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}+{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}={\sqrt {2}}\left({\sqrt {5}}+1\right)}
d)
3
−
1
3
+
1
=
9
−
5
3
9
+
5
3
3
{\displaystyle {\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}={\sqrt[{3}]{\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}}}
e)
1
+
3
2
2
3
=
2
+
3
20
+
12
3
3
{\displaystyle {\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}={\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}}
f)
38
+
17
5
3
=
9
+
4
5
{\displaystyle {\sqrt[{3}]{38+17{\sqrt {5}}}}={\sqrt {9+4{\sqrt {5}}}}}
a) Corrigé.
(
11
5
−
3
)
2
−
(
5
−
2
5
2
−
5
)
2
=
(
11
(
5
+
3
)
(
5
−
3
)
(
5
+
3
)
)
2
−
(
(
5
−
2
5
)
(
2
+
5
)
(
2
−
5
)
(
2
+
5
)
)
2
=
(
11
(
5
+
3
)
22
)
2
−
(
10
+
5
5
−
4
5
−
2
×
5
−
1
)
2
=
(
5
+
3
2
)
2
−
(
−
5
)
2
=
25
+
10
3
+
3
4
−
20
4
=
8
+
10
3
4
=
4
+
5
3
2
{\displaystyle {\begin{aligned}\left({\frac {11}{5-{\sqrt {3}}}}\right)^{2}-\left({\frac {5-2{\sqrt {5}}}{2-{\sqrt {5}}}}\right)^{2}&=\left({\frac {11(5+{\sqrt {3}})}{(5-{\sqrt {3}})(5+{\sqrt {3}})}}\right)^{2}-\left({\frac {(5-2{\sqrt {5}})(2+{\sqrt {5}})}{(2-{\sqrt {5}})(2+{\sqrt {5}})}}\right)^{2}\\&=\left({\frac {11(5+{\sqrt {3}})}{22}}\right)^{2}-\left({\frac {10+5{\sqrt {5}}-4{\sqrt {5}}-2\times 5}{-1}}\right)^{2}\\&=\left({\frac {5+{\sqrt {3}}}{2}}\right)^{2}-\left(-{\sqrt {5}}\right)^{2}\\&={\frac {25+10{\sqrt {3}}+3}{4}}-{\frac {20}{4}}\\&={\frac {8+10{\sqrt {3}}}{4}}\\&={\frac {4+5{\sqrt {3}}}{2}}\\\end{aligned}}}
On est donc ramené à vérifier que :
4
+
5
3
2
=
91
4
+
10
3
{\displaystyle {\frac {4+5{\sqrt {3}}}{2}}={\sqrt {{\frac {91}{4}}+10{\sqrt {3}}}}}
Ce qui se fait aisément en remarquant que :
(
4
+
5
3
2
)
2
=
16
+
2
×
4
×
5
3
+
75
4
=
91
+
40
3
4
=
91
4
+
10
3
{\displaystyle \left({\frac {4+5{\sqrt {3}}}{2}}\right)^{2}={\frac {16+2\times 4\times 5{\sqrt {3}}+75}{4}}={\frac {91+40{\sqrt {3}}}{4}}={\frac {91}{4}}+10{\sqrt {3}}}
b) Corrigé.
La relation est équivalente à :
4
(
9
+
65
)
(
9
−
65
)
=
(
19
+
3
)
(
19
−
3
)
{\displaystyle 4{\sqrt {(9+{\sqrt {65}})(9-{\sqrt {65}})}}=({\sqrt {19}}+{\sqrt {3}})({\sqrt {19}}-{\sqrt {3}})}
qui s'écrit :
4
81
−
65
=
19
−
3
{\displaystyle 4{\sqrt {81-65}}=19-3}
soit
16
=
4
{\displaystyle {\sqrt {16}}=4}
Ce qui est vrai.
c) Corrigé.
Comme les deux membres sont positifs, montrons que le carré des deux membres sont égaux.
Le carré du premier membre est :
(
8
+
2
10
+
2
5
+
8
−
2
10
+
2
5
)
2
=
8
+
2
10
+
2
5
+
2
8
+
2
10
+
2
5
8
−
2
10
+
2
5
+
8
−
2
10
+
2
5
=
16
+
2
64
−
4
(
10
+
2
5
)
=
16
+
2
24
−
8
5
=
16
+
4
6
−
2
5
{\displaystyle {\begin{aligned}\left({\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}+{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}\right)^{2}&=8+2{\sqrt {10+2{\sqrt {5}}}}+2{\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}+8-2{\sqrt {10+2{\sqrt {5}}}}\\&=16+2{\sqrt {64-4(10+2{\sqrt {5}})}}\\&=16+2{\sqrt {24-8{\sqrt {5}}}}\\&=16+4{\sqrt {6-2{\sqrt {5}}}}\\\end{aligned}}}
Le carré du deuxième membre est :
(
2
(
5
+
1
)
)
2
=
2
(
6
+
2
5
)
{\displaystyle \left({\sqrt {2}}\left({\sqrt {5}}+1\right)\right)^{2}=2\left(6+2{\sqrt {5}}\right)}
On s'est donc ramené à établir la relation plus simple :
16
+
4
6
−
2
5
=
2
(
6
+
2
5
)
{\displaystyle 16+4{\sqrt {6-2{\sqrt {5}}}}=2\left(6+2{\sqrt {5}}\right)}
qui s'écrit encore plus simplement :
6
−
2
5
=
5
−
1
{\displaystyle {\sqrt {6-2{\sqrt {5}}}}={\sqrt {5}}-1}
Qui est vraie puisque :
(
5
−
1
)
2
=
6
−
2
5
{\displaystyle \left({\sqrt {5}}-1\right)^{2}=6-2{\sqrt {5}}}
d) Corrigé.
Il nous suffit de montrer que :
(
3
−
1
3
+
1
)
3
=
9
−
5
3
9
+
5
3
{\displaystyle \left({\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}\right)^{3}={\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}}
Effectivement, on a bien :
(
3
−
1
3
+
1
)
3
=
(
3
)
3
−
3
(
3
)
2
+
3
3
−
1
3
(
3
)
3
+
3
(
3
)
2
+
3
3
+
1
3
=
3
3
−
9
+
3
3
−
1
3
3
+
9
+
3
3
+
1
=
6
3
−
10
6
3
+
10
=
3
3
−
5
3
3
+
5
=
3
(
3
3
−
5
)
3
(
3
3
+
5
)
=
9
−
5
3
9
+
5
3
{\displaystyle {\begin{aligned}\left({\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}\right)^{3}&={\frac {({\sqrt {3}})^{3}-3({\sqrt {3}})^{2}+3{\sqrt {3}}-1^{3}}{({\sqrt {3}})^{3}+3({\sqrt {3}})^{2}+3{\sqrt {3}}+1^{3}}}\\&={\frac {3{\sqrt {3}}-9+3{\sqrt {3}}-1}{3{\sqrt {3}}+9+3{\sqrt {3}}+1}}\\&={\frac {6{\sqrt {3}}-10}{6{\sqrt {3}}+10}}\\&={\frac {3{\sqrt {3}}-5}{3{\sqrt {3}}+5}}\\&={\frac {{\sqrt {3}}\left(3{\sqrt {3}}-5\right)}{{\sqrt {3}}\left(3{\sqrt {3}}+5\right)}}\\&={\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}\end{aligned}}}
e) Corrigé.
Calculons le cube des deux membres :
Pour le premier membre :
(
1
+
3
2
2
3
)
3
=
1
3
+
3
3
+
3
(
3
)
2
+
(
3
)
3
(
2
2
3
)
3
=
1
+
3
3
+
9
+
3
3
16
=
10
+
6
3
16
=
5
+
3
3
8
{\displaystyle {\begin{aligned}\left({\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}\right)^{3}&={\frac {1^{3}+3{\sqrt {3}}+3({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{(2{\sqrt[{3}]{2}})^{3}}}\\&={\frac {1+3{\sqrt {3}}+9+3{\sqrt {3}}}{16}}\\&={\frac {10+6{\sqrt {3}}}{16}}\\&={\frac {5+3{\sqrt {3}}}{8}}\\\end{aligned}}}
Pour le deuxième membre :
(
2
+
3
20
+
12
3
3
)
3
=
2
3
+
3
×
2
2
3
+
3
×
2
(
3
)
2
+
(
3
)
3
20
+
12
3
=
8
+
12
3
+
18
+
3
3
20
+
12
3
=
26
+
15
3
20
+
12
3
=
(
26
+
15
3
)
(
20
−
12
3
)
(
20
+
12
3
)
(
20
−
12
3
)
=
520
−
312
3
+
300
3
−
540
400
−
432
=
−
20
−
12
3
−
32
=
5
+
3
3
8
{\displaystyle {\begin{aligned}\left({\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}\right)^{3}&={\frac {2^{3}+3\times 2^{2}{\sqrt {3}}+3\times 2({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{20+12{\sqrt {3}}}}\\&={\frac {8+12{\sqrt {3}}+18+3{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {26+15{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {(26+15{\sqrt {3}})(20-12{\sqrt {3}})}{(20+12{\sqrt {3}})(20-12{\sqrt {3}})}}\\&={\frac {520-312{\sqrt {3}}+300{\sqrt {3}}-540}{400-432}}\\&={\frac {-20-12{\sqrt {3}}}{-32}}\\&={\frac {5+3{\sqrt {3}}}{8}}\end{aligned}}}
En élevant les deux membres au cube, nous obtenons le même résultat. On a donc bien :
1
+
3
2
2
3
=
2
+
3
20
+
12
3
3
{\displaystyle {\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}={\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}}
f) Corrigé.
Pour établir cette égalité, nous comparerons les deux membres élevées à la puissance 6 .
Pour le premier membre, on a :
(
38
+
17
5
3
)
6
=
(
38
+
17
5
)
2
=
38
2
+
2
×
38
×
17
5
+
(
17
5
)
2
=
2889
+
1292
5
{\displaystyle {\begin{aligned}\left({\sqrt[{3}]{38+17{\sqrt {5}}}}\right)^{6}&=\left(38+17{\sqrt {5}}\right)^{2}\\&=38^{2}+2\times 38\times 17{\sqrt {5}}+\left(17{\sqrt {5}}\right)^{2}\\&=2889+1292{\sqrt {5}}\end{aligned}}}
Pour le deuxième membre, on a :
(
9
+
4
5
)
6
=
(
9
+
4
5
)
3
=
9
3
+
3
×
9
2
×
4
5
+
3
×
9
×
(
4
5
)
2
+
(
4
5
)
3
=
2889
+
1292
5
{\displaystyle {\begin{aligned}\left({\sqrt {9+4{\sqrt {5}}}}\right)^{6}&=\left(9+4{\sqrt {5}}\right)^{3}\\&=9^{3}+3\times 9^{2}\times 4{\sqrt {5}}+3\times 9\times \left(4{\sqrt {5}}\right)^{2}+\left(4{\sqrt {5}}\right)^{3}\\&=2889+1292{\sqrt {5}}\end{aligned}}}
Les deux membres à la puissance 6 étant égaux, on a bien :
38
+
17
5
3
=
9
+
4
5
{\displaystyle {\sqrt[{3}]{38+17{\sqrt {5}}}}={\sqrt {9+4{\sqrt {5}}}}}
Simplifier les expressions suivantes :
a)
5
+
2
6
3
+
2
2
{\displaystyle {\sqrt {\frac {5+2{\sqrt {6}}}{3+2{\sqrt {2}}}}}}
b)
7
+
2
10
5
+
2
6
{\displaystyle {\sqrt {\frac {7+2{\sqrt {10}}}{5+2{\sqrt {6}}}}}}
c)
5
−
6
13
+
5
6
{\displaystyle {\sqrt {\frac {5-{\sqrt {6}}}{13+5{\sqrt {6}}}}}}
d)
4
+
7
3
+
5
{\displaystyle {\sqrt {\frac {4+{\sqrt {7}}}{3+{\sqrt {5}}}}}}
e)
2
+
3
2
+
2
+
3
+
2
−
3
2
−
2
−
3
{\displaystyle {\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {2+{\sqrt {3}}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {2-{\sqrt {3}}}}}}}
f)
3
−
2
2
17
−
12
2
−
3
+
2
2
17
+
12
2
{\displaystyle {\frac {\sqrt {3-2{\sqrt {2}}}}{\sqrt {17-12{\sqrt {2}}}}}-{\frac {\sqrt {3+2{\sqrt {2}}}}{\sqrt {17+12{\sqrt {2}}}}}}
h)
8
4
+
2
−
1
−
8
4
−
2
−
1
8
4
−
2
+
1
{\displaystyle {\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}}
a) Corrigé.
5
+
2
6
3
+
2
2
=
3
+
2
2
3
+
2
2
+
2
2
+
1
=
(
3
)
2
+
2
2
3
+
(
2
)
2
(
2
)
2
+
2
2
+
1
=
(
3
+
2
)
2
(
2
+
1
)
2
=
3
+
2
2
+
1
=
(
3
+
2
)
(
2
−
1
)
(
2
+
1
)
(
2
−
1
)
=
(
3
+
2
)
(
2
−
1
)
{\displaystyle {\begin{aligned}{\sqrt {\frac {5+2{\sqrt {6}}}{3+2{\sqrt {2}}}}}&={\sqrt {\frac {3+2{\sqrt {2}}{\sqrt {3}}+2}{2+2{\sqrt {2}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {3}}\right)^{2}+2{\sqrt {2}}{\sqrt {3}}+\left({\sqrt {2}}\right)^{2}}{\left({\sqrt {2}}\right)^{2}+2{\sqrt {2}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {3}}+{\sqrt {2}}\right)^{2}}{\left({\sqrt {2}}+1\right)^{2}}}}\\&={\frac {{\sqrt {3}}+{\sqrt {2}}}{{\sqrt {2}}+1}}\\&={\frac {({\sqrt {3}}+{\sqrt {2}})({\sqrt {2}}-1)}{({\sqrt {2}}+1)({\sqrt {2}}-1)}}\\&=({\sqrt {3}}+{\sqrt {2}})({\sqrt {2}}-1)\end{aligned}}}
b) Corrigé.
7
+
2
10
5
+
2
6
=
5
+
2
2
5
+
2
3
+
2
2
3
+
2
=
(
5
)
2
+
2
2
5
+
(
2
)
2
(
3
)
2
+
2
2
3
+
(
2
)
2
=
(
5
+
2
)
2
(
3
+
2
)
2
=
5
+
2
3
+
2
=
(
5
+
2
)
(
3
−
2
)
(
3
+
2
)
(
3
−
2
)
=
(
5
+
2
)
(
3
−
2
)
3
−
2
=
(
5
+
2
)
(
3
−
2
)
{\displaystyle {\begin{aligned}{\sqrt {\frac {7+2{\sqrt {10}}}{5+2{\sqrt {6}}}}}&={\sqrt {\frac {5+2{\sqrt {2}}{\sqrt {5}}+2}{3+2{\sqrt {2}}{\sqrt {3}}+2}}}\\&={\sqrt {\frac {\left({\sqrt {5}}\right)^{2}+2{\sqrt {2}}{\sqrt {5}}+\left({\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}\right)^{2}+2{\sqrt {2}}{\sqrt {3}}+\left({\sqrt {2}}\right)^{2}}}}\\&={\sqrt {\frac {\left({\sqrt {5}}+{\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}+{\sqrt {2}}\right)^{2}}}}\\&={\frac {{\sqrt {5}}+{\sqrt {2}}}{{\sqrt {3}}+{\sqrt {2}}}}\\&={\frac {({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}{({\sqrt {3}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}}\\&={\frac {({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}{3-2}}\\&=({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})\end{aligned}}}
c) Corrigé.
5
−
6
13
+
5
6
=
(
5
−
6
)
(
13
−
5
6
)
(
13
+
5
6
)
(
13
−
5
6
)
=
65
−
25
6
−
13
6
+
30
169
−
150
=
95
−
38
6
19
=
5
−
2
6
=
2
−
2
6
+
3
=
(
2
+
3
)
2
=
2
+
3
{\displaystyle {\begin{aligned}{\sqrt {\frac {5-{\sqrt {6}}}{13+5{\sqrt {6}}}}}&={\sqrt {\frac {(5-{\sqrt {6}})(13-5{\sqrt {6}})}{(13+5{\sqrt {6}})(13-5{\sqrt {6}})}}}\\&={\sqrt {\frac {65-25{\sqrt {6}}-13{\sqrt {6}}+30}{169-150}}}\\&={\sqrt {\frac {95-38{\sqrt {6}}}{19}}}\\&={\sqrt {5-2{\sqrt {6}}}}\\&={\sqrt {2-2{\sqrt {6}}+3}}\\&={\sqrt {\left({\sqrt {2}}+{\sqrt {3}}\right)^{2}}}\\&={\sqrt {2}}+{\sqrt {3}}\end{aligned}}}
d) Corrigé.
4
+
7
3
+
5
=
8
+
2
7
6
+
2
5
=
7
+
2
7
+
1
5
+
2
5
+
1
=
(
7
+
1
)
2
(
5
+
1
)
2
=
7
+
1
5
+
1
=
(
7
+
1
)
(
5
−
1
)
(
5
+
1
)
(
5
−
1
)
=
35
−
7
+
5
−
1
4
{\displaystyle {\begin{aligned}{\sqrt {\frac {4+{\sqrt {7}}}{3+{\sqrt {5}}}}}&={\sqrt {\frac {8+2{\sqrt {7}}}{6+2{\sqrt {5}}}}}\\&={\sqrt {\frac {7+2{\sqrt {7}}+1}{5+2{\sqrt {5}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {7}}+1\right)^{2}}{\left({\sqrt {5}}+1\right)^{2}}}}\\&={\frac {{\sqrt {7}}+1}{{\sqrt {5}}+1}}\\&={\frac {({\sqrt {7}}+1)({\sqrt {5}}-1)}{({\sqrt {5}}+1)({\sqrt {5}}-1)}}\\&={\frac {{\sqrt {35}}-{\sqrt {7}}+{\sqrt {5}}-1}{4}}\end{aligned}}}
e) Corrigé.
2
+
3
2
+
2
+
3
+
2
−
3
2
−
2
−
3
=
2
+
3
2
+
3
+
2
3
+
1
2
+
2
−
3
2
−
3
−
2
3
+
1
2
=
2
+
3
2
+
(
3
+
1
)
2
2
+
2
−
3
2
−
(
3
−
1
)
2
2
=
2
+
3
2
+
3
+
1
2
+
2
−
3
2
−
3
−
1
2
=
2
2
+
6
2
+
3
+
1
+
2
2
−
6
2
−
3
+
1
=
2
2
+
6
3
+
3
+
2
2
−
6
3
−
3
=
(
2
2
+
6
)
(
3
−
3
)
+
(
2
2
−
6
)
(
3
+
3
)
(
3
+
3
)
(
3
−
3
)
=
6
2
−
2
6
+
3
6
−
3
2
+
6
2
+
2
6
−
3
6
−
3
2
9
−
3
=
6
2
6
=
2
{\displaystyle {\begin{aligned}{\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {2+{\sqrt {3}}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {2-{\sqrt {3}}}}}}&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {\frac {3+2{\sqrt {3}}+1}{2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {\frac {3-2{\sqrt {3}}+1}{2}}}}}\\&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\frac {\sqrt {\left({\sqrt {3}}+1\right)^{2}}}{\sqrt {2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\frac {\sqrt {\left({\sqrt {3}}-1\right)^{2}}}{\sqrt {2}}}}}\\&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\frac {{\sqrt {3}}+1}{\sqrt {2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\frac {{\sqrt {3}}-1}{\sqrt {2}}}}}\\&={\frac {2{\sqrt {2}}+{\sqrt {6}}}{2+{\sqrt {3}}+1}}+{\frac {2{\sqrt {2}}-{\sqrt {6}}}{2-{\sqrt {3}}+1}}\\&={\frac {2{\sqrt {2}}+{\sqrt {6}}}{3+{\sqrt {3}}}}+{\frac {2{\sqrt {2}}-{\sqrt {6}}}{3-{\sqrt {3}}}}\\&={\frac {(2{\sqrt {2}}+{\sqrt {6}})(3-{\sqrt {3}})+(2{\sqrt {2}}-{\sqrt {6}})(3+{\sqrt {3}})}{(3+{\sqrt {3}})(3-{\sqrt {3}})}}\\&={\frac {6{\sqrt {2}}-2{\sqrt {6}}+3{\sqrt {6}}-3{\sqrt {2}}+6{\sqrt {2}}+2{\sqrt {6}}-3{\sqrt {6}}-3{\sqrt {2}}}{9-3}}\\&={\frac {6{\sqrt {2}}}{6}}\\&={\sqrt {2}}\end{aligned}}}
f) Corrigé.
3
−
2
2
17
−
12
2
−
3
+
2
2
17
+
12
2
=
2
−
2
2
+
1
9
−
2
×
3
8
+
8
−
2
+
2
2
+
1
9
+
2
×
3
8
+
8
=
(
2
−
1
)
2
(
3
−
8
)
2
−
(
2
+
1
)
2
(
3
+
8
)
2
=
2
−
1
3
−
8
−
2
+
1
3
+
8
=
(
2
−
1
)
(
3
+
8
)
−
(
2
+
1
)
(
3
−
8
)
(
3
−
8
)
(
3
+
8
)
=
3
2
+
4
−
3
−
8
−
3
2
+
4
−
3
+
8
9
−
8
=
2
{\displaystyle {\begin{aligned}{\frac {\sqrt {3-2{\sqrt {2}}}}{\sqrt {17-12{\sqrt {2}}}}}-{\frac {\sqrt {3+2{\sqrt {2}}}}{\sqrt {17+12{\sqrt {2}}}}}&={\frac {\sqrt {2-2{\sqrt {2}}+1}}{\sqrt {9-2\times 3{\sqrt {8}}+8}}}-{\frac {\sqrt {2+2{\sqrt {2}}+1}}{\sqrt {9+2\times 3{\sqrt {8}}+8}}}\\&={\frac {\sqrt {\left({\sqrt {2}}-1\right)^{2}}}{\sqrt {\left(3-{\sqrt {8}}\right)^{2}}}}-{\frac {\sqrt {\left({\sqrt {2}}+1\right)^{2}}}{\sqrt {\left(3+{\sqrt {8}}\right)^{2}}}}\\&={\frac {{\sqrt {2}}-1}{3-{\sqrt {8}}}}-{\frac {{\sqrt {2}}+1}{3+{\sqrt {8}}}}\\&={\frac {({\sqrt {2}}-1)(3+{\sqrt {8}})-({\sqrt {2}}+1)(3-{\sqrt {8}})}{(3-{\sqrt {8}})(3+{\sqrt {8}})}}\\&={\frac {3{\sqrt {2}}+4-3-{\sqrt {8}}-3{\sqrt {2}}+4-3+{\sqrt {8}}}{9-8}}\\&=2\end{aligned}}}
h) Corrigé.
8
4
+
2
−
1
−
8
4
−
2
−
1
8
4
−
2
+
1
=
(
8
4
+
2
−
1
−
8
4
−
2
−
1
)
2
8
4
−
2
+
1
=
8
4
+
2
−
1
−
2
8
4
+
2
−
1
8
4
−
2
−
1
+
8
4
−
2
−
1
8
4
−
2
+
1
=
2
8
4
−
2
(
8
4
+
2
−
1
)
(
8
4
−
2
−
1
)
8
4
−
2
+
1
=
2
8
4
−
8
−
(
2
−
1
)
8
4
−
2
+
1
=
2
8
4
−
2
2
−
2
+
1
8
4
−
2
+
1
=
2
8
4
−
2
+
1
8
4
−
2
+
1
=
2
{\displaystyle {\begin{aligned}{\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}&={\frac {\sqrt {\left({\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}\right)^{2}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}-2{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}+{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {2{\sqrt[{4}]{8}}-2{\sqrt {\left({\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}\right)\left({\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {8}}-\left({\sqrt {2}}-1\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {2{\sqrt {2}}-{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}\end{aligned}}}
Simplifier les expressions littérales suivantes :
a)
a
+
a
2
−
x
2
a
−
a
2
−
x
2
+
a
−
a
2
−
x
2
a
+
a
2
−
x
2
{\displaystyle {\frac {a+{\sqrt {a^{2}-x^{2}}}}{a-{\sqrt {a^{2}-x^{2}}}}}+{\frac {a-{\sqrt {a^{2}-x^{2}}}}{a+{\sqrt {a^{2}-x^{2}}}}}}
b)
x
+
x
2
−
1
x
−
x
2
−
1
−
x
−
x
2
−
1
x
+
x
2
−
1
{\displaystyle {\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}-{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}}
c)
1
+
x
2
1
−
x
2
+
1
−
x
2
1
+
x
2
−
2
1
−
x
4
{\displaystyle {\frac {\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}+{\frac {\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}-{\frac {2}{\sqrt {1-x^{4}}}}}
d)
(
x
+
x
2
−
a
2
)
4
−
a
4
4
(
x
+
x
2
−
a
2
)
2
{\displaystyle {\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}-a^{4}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}}}}
e)
x
2
−
x
−
2
+
(
x
−
1
)
x
2
−
4
x
2
+
x
−
2
+
(
x
+
1
)
x
2
−
4
{\displaystyle {\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}}
a) Corrigé.
a
+
a
2
−
x
2
a
−
a
2
−
x
2
+
a
−
a
2
−
x
2
a
+
a
2
−
x
2
=
(
a
+
a
2
−
x
2
)
(
a
+
a
2
−
x
2
)
(
a
−
a
2
−
x
2
)
(
a
+
a
2
−
x
2
)
+
(
a
−
a
2
−
x
2
)
(
a
−
a
2
−
x
2
)
(
a
+
a
2
−
x
2
)
(
a
−
a
2
−
x
2
)
=
(
a
+
a
2
−
x
2
)
2
a
2
−
(
a
2
−
x
2
)
+
(
a
−
a
2
−
x
2
)
2
a
2
−
(
a
2
−
x
2
)
=
(
a
+
a
2
−
x
2
)
2
x
2
+
(
a
−
a
2
−
x
2
)
2
x
2
=
a
2
+
2
a
a
2
−
x
2
+
a
2
−
x
2
+
a
2
−
2
a
a
2
−
x
2
+
a
2
−
x
2
x
2
=
4
a
2
−
2
x
2
x
2
=
2
(
2
a
2
−
x
2
)
x
2
{\displaystyle {\begin{aligned}{\frac {a+{\sqrt {a^{2}-x^{2}}}}{a-{\sqrt {a^{2}-x^{2}}}}}+{\frac {a-{\sqrt {a^{2}-x^{2}}}}{a+{\sqrt {a^{2}-x^{2}}}}}&={\frac {(a+{\sqrt {a^{2}-x^{2}}})(a+{\sqrt {a^{2}-x^{2}}})}{(a-{\sqrt {a^{2}-x^{2}}})(a+{\sqrt {a^{2}-x^{2}}})}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})(a-{\sqrt {a^{2}-x^{2}}})}{(a+{\sqrt {a^{2}-x^{2}}})(a-{\sqrt {a^{2}-x^{2}}})}}\\&={\frac {(a+{\sqrt {a^{2}-x^{2}}})^{2}}{a^{2}-(a^{2}-x^{2})}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})^{2}}{a^{2}-(a^{2}-x^{2})}}\\&={\frac {(a+{\sqrt {a^{2}-x^{2}}})^{2}}{x^{2}}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})^{2}}{x^{2}}}\\&={\frac {a^{2}+2a{\sqrt {a^{2}-x^{2}}}+a^{2}-x^{2}+a^{2}-2a{\sqrt {a^{2}-x^{2}}}+a^{2}-x^{2}}{x^{2}}}\\&={\frac {4a^{2}-2x^{2}}{x^{2}}}\\&={\frac {2(2a^{2}-x^{2})}{x^{2}}}\end{aligned}}}
b) Corrigé.
x
+
x
2
−
1
x
−
x
2
−
1
−
x
−
x
2
−
1
x
+
x
2
−
1
=
(
x
+
x
2
−
1
)
(
x
+
x
2
−
1
)
(
x
−
x
2
−
1
)
(
x
+
x
2
−
1
)
−
(
x
−
x
2
−
1
)
(
x
−
x
2
−
1
)
(
x
+
x
2
−
1
)
(
x
−
x
2
−
1
)
=
(
x
+
x
2
−
1
)
2
x
2
−
(
x
2
−
1
)
−
(
x
−
x
2
−
1
)
2
x
2
−
(
x
2
−
1
)
=
(
x
+
x
2
−
1
)
2
−
(
x
−
x
2
−
1
)
2
=
(
x
2
+
2
x
x
2
−
1
+
x
2
−
1
)
−
(
x
2
−
2
x
x
2
−
1
+
x
2
−
1
)
=
4
x
x
2
−
1
{\displaystyle {\begin{aligned}{\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}-{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}&={\frac {(x+{\sqrt {x^{2}-1}})(x+{\sqrt {x^{2}-1}})}{(x-{\sqrt {x^{2}-1}})(x+{\sqrt {x^{2}-1}})}}-{\frac {(x-{\sqrt {x^{2}-1}})(x-{\sqrt {x^{2}-1}})}{(x+{\sqrt {x^{2}-1}})(x-{\sqrt {x^{2}-1}})}}\\&={\frac {(x+{\sqrt {x^{2}-1}})^{2}}{x^{2}-(x^{2}-1)}}-{\frac {(x-{\sqrt {x^{2}-1}})^{2}}{x^{2}-(x^{2}-1)}}\\&=(x+{\sqrt {x^{2}-1}})^{2}-(x-{\sqrt {x^{2}-1}})^{2}\\&=\left(x^{2}+2x{\sqrt {x^{2}-1}}+x^{2}-1\right)-\left(x^{2}-2x{\sqrt {x^{2}-1}}+x^{2}-1\right)\\&=4x{\sqrt {x^{2}-1}}\end{aligned}}}
c) Corrigé.
1
+
x
2
1
−
x
2
+
1
−
x
2
1
+
x
2
−
2
1
−
x
4
=
1
+
x
2
1
+
x
2
1
−
x
2
1
+
x
2
+
1
−
x
2
1
−
x
2
1
+
x
2
1
−
x
2
−
2
1
−
x
4
=
1
+
x
2
1
−
x
4
+
1
−
x
2
1
−
x
4
−
2
1
−
x
4
=
1
+
x
2
+
1
−
x
2
−
2
1
−
x
4
=
0
{\displaystyle {\begin{aligned}{\frac {\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}+{\frac {\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}-{\frac {2}{\sqrt {1-x^{4}}}}&={\frac {{\sqrt {1+x^{2}}}{\sqrt {1+x^{2}}}}{{\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}}+{\frac {{\sqrt {1-x^{2}}}{\sqrt {1-x^{2}}}}{{\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}}-{\frac {2}{\sqrt {1-x^{4}}}}\\&={\frac {1+x^{2}}{\sqrt {1-x^{4}}}}+{\frac {1-x^{2}}{\sqrt {1-x^{4}}}}-{\frac {2}{\sqrt {1-x^{4}}}}\\&={\frac {1+x^{2}+1-x^{2}-2}{\sqrt {1-x^{4}}}}\\&=0\end{aligned}}}
d) Corrigé.
(
x
+
x
2
−
a
2
)
4
−
a
4
4
(
x
+
x
2
−
a
2
)
2
=
(
x
+
x
2
−
a
2
)
4
(
x
−
x
2
−
a
2
)
2
−
a
4
(
x
−
x
2
−
a
2
)
2
4
(
x
+
x
2
−
a
2
)
2
(
x
−
x
2
−
a
2
)
2
=
(
x
+
x
2
−
a
2
)
2
(
x
2
−
(
x
2
−
a
2
)
)
2
−
a
4
(
x
−
x
2
−
a
2
)
2
4
(
x
2
−
(
x
2
−
a
2
)
)
2
=
a
4
(
x
+
x
2
−
a
2
)
2
−
a
4
(
x
−
x
2
−
a
2
)
2
4
a
4
=
(
x
+
x
2
−
a
2
)
2
−
(
x
−
x
2
−
a
2
)
2
4
=
(
x
2
+
2
x
x
2
−
a
2
+
x
2
−
a
2
)
−
(
x
2
−
2
x
x
2
−
a
2
+
x
2
−
a
2
)
4
=
x
x
2
−
a
2
{\displaystyle {\begin{aligned}{\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}-a^{4}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}}}&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}}\\&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}\left(x^{2}-(x^{2}-a^{2})\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4\left(x^{2}-(x^{2}-a^{2})\right)^{2}}}\\&={\frac {a^{4}\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4a^{4}}}\\&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}-\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4}}\\&={\frac {\left(x^{2}+2x{\sqrt {x^{2}-a^{2}}}+x^{2}-a^{2}\right)-\left(x^{2}-2x{\sqrt {x^{2}-a^{2}}}+x^{2}-a^{2}\right)}{4}}\\&=x{\sqrt {x^{2}-a^{2}}}\end{aligned}}}
e) Corrigé.
x
2
−
x
−
2
+
(
x
−
1
)
x
2
−
4
x
2
+
x
−
2
+
(
x
+
1
)
x
2
−
4
=
(
x
+
1
)
(
x
−
2
)
+
(
x
−
1
)
x
2
−
4
(
x
−
1
)
(
x
+
2
)
+
(
x
+
1
)
x
2
−
4
{\displaystyle {\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}={\frac {(x+1)(x-2)+(x-1){\sqrt {x^{2}-4}}}{(x-1)(x+2)+(x+1){\sqrt {x^{2}-4}}}}}
L'expression sous le radical n'est définie que si
x
⩾
2
{\displaystyle x\geqslant 2}
x
⩽
−
2
{\displaystyle x\leqslant -2}
Nous devons aussi exclure la valeur
x
=
−
2
{\displaystyle x=-2}
Premier cas :
x
⩾
2
{\displaystyle x\geqslant 2}
x
2
−
x
−
2
+
(
x
−
1
)
x
2
−
4
x
2
+
x
−
2
+
(
x
+
1
)
x
2
−
4
=
(
x
+
1
)
(
x
−
2
)
+
(
x
−
1
)
(
x
−
2
)
(
x
+
2
)
(
x
−
1
)
(
x
+
2
)
+
(
x
+
1
)
(
x
−
2
)
(
x
+
2
)
=
(
x
+
1
)
x
−
2
x
−
2
+
(
x
−
1
)
x
−
2
x
+
2
(
x
−
1
)
x
+
2
x
+
2
+
(
x
+
1
)
x
−
2
x
+
2
=
x
−
2
[
(
x
+
1
)
x
−
2
+
(
x
−
1
)
x
+
2
]
x
+
2
[
(
x
−
1
)
x
+
2
+
(
x
+
1
)
x
−
2
]
=
x
−
2
x
+
2
=
x
−
2
x
+
2
{\displaystyle {\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(x-2)(x+2)}}}{(x-1)(x+2)+(x+1){\sqrt {(x-2)(x+2)}}}}\\&={\frac {(x+1){\sqrt {x-2}}{\sqrt {x-2}}+(x-1){\sqrt {x-2}}{\sqrt {x+2}}}{(x-1){\sqrt {x+2}}{\sqrt {x+2}}+(x+1){\sqrt {x-2}}{\sqrt {x+2}}}}\\&={\frac {{\sqrt {x-2}}\left[(x+1){\sqrt {x-2}}+(x-1){\sqrt {x+2}}\right]}{{\sqrt {x+2}}\left[(x-1){\sqrt {x+2}}+(x+1){\sqrt {x-2}}\right]}}\\&={\frac {\sqrt {x-2}}{\sqrt {x+2}}}\\&={\sqrt {\frac {x-2}{x+2}}}\end{aligned}}}
Deuxième cas :
x
<
−
2
{\displaystyle x<-2}
x
2
−
x
−
2
+
(
x
−
1
)
x
2
−
4
x
2
+
x
−
2
+
(
x
+
1
)
x
2
−
4
=
(
x
+
1
)
(
x
−
2
)
+
(
x
−
1
)
(
2
−
x
)
(
−
2
−
x
)
(
x
−
1
)
(
x
+
2
)
+
(
x
+
1
)
(
2
−
x
)
(
−
2
−
x
)
=
−
(
x
+
1
)
2
−
x
2
−
x
+
(
x
−
1
)
2
−
x
−
2
−
x
−
(
x
−
1
)
−
2
−
x
−
2
−
x
+
(
x
+
1
)
2
−
x
−
2
−
x
=
2
−
x
[
−
(
x
+
1
)
2
−
x
+
(
x
−
1
)
−
2
−
x
]
−
2
−
x
[
−
(
x
−
1
)
−
2
−
x
+
(
x
+
1
)
2
−
x
]
=
−
2
−
x
−
2
−
x
=
−
2
−
x
−
2
−
x
=
−
x
−
2
x
+
2
{\displaystyle {\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(2-x)(-2-x)}}}{(x-1)(x+2)+(x+1){\sqrt {(2-x)(-2-x)}}}}\\&={\frac {-(x+1){\sqrt {2-x}}{\sqrt {2-x}}+(x-1){\sqrt {2-x}}{\sqrt {-2-x}}}{-(x-1){\sqrt {-2-x}}{\sqrt {-2-x}}+(x+1){\sqrt {2-x}}{\sqrt {-2-x}}}}\\&={\frac {{\sqrt {2-x}}\left[-(x+1){\sqrt {2-x}}+(x-1){\sqrt {-2-x}}\right]}{{\sqrt {-2-x}}\left[-(x-1){\sqrt {-2-x}}+(x+1){\sqrt {2-x}}\right]}}\\&=-{\frac {\sqrt {2-x}}{\sqrt {-2-x}}}\\&=-{\sqrt {\frac {2-x}{-2-x}}}\\&=-{\sqrt {\frac {x-2}{x+2}}}\end{aligned}}}
On suppose que :
a
d
=
b
c
{\displaystyle ad=bc}
en déduire que :
a
b
+
c
d
=
(
a
+
c
)
(
b
+
d
)
{\displaystyle {\sqrt {ab}}+{\sqrt {cd}}={\sqrt {(a+c)(b+d)}}}
corrigé
Si
a
=
0
{\displaystyle a=0}
c
=
0
{\displaystyle c=0}
Si
a
{\displaystyle a}
c
{\displaystyle c}
(
a
d
=
b
c
)
⟺
(
b
a
=
d
c
)
{\displaystyle (ad=bc)\Longleftrightarrow \left({\frac {b}{a}}={\frac {d}{c}}\right)}
posons alors :
k
=
b
a
=
d
c
{\displaystyle k={\frac {b}{a}}={\frac {d}{c}}}
on alors :
{
b
=
k
a
d
=
k
c
{\displaystyle {\begin{cases}b=ka\\d=kc\end{cases}}}
b
+
d
=
b
+
d
{\displaystyle b+d=b+d}
qui peut s'écrire :
b
2
+
d
2
=
(
b
+
d
)
2
{\displaystyle {\sqrt {b^{2}}}+{\sqrt {d^{2}}}={\sqrt {(b+d)^{2}}}}
soit :
k
a
b
+
k
c
d
=
(
k
a
+
k
c
)
(
b
+
d
)
{\displaystyle {\sqrt {kab}}+{\sqrt {kcd}}={\sqrt {(ka+kc)(b+d)}}}
et en simplifiant par
k
{\displaystyle {\sqrt {k}}}
a
b
+
c
d
=
(
a
+
c
)
(
b
+
d
)
{\displaystyle {\sqrt {ab}}+{\sqrt {cd}}={\sqrt {(a+c)(b+d)}}}
Pour tout
a
>
0
,
b
>
0
,
a
>
b
,
{\displaystyle a>0,\,b>0,\,a>b,}
2
(
2
a
+
a
2
−
b
2
)
a
−
a
2
−
b
2
=
(
a
+
b
)
3
−
(
a
−
b
)
3
{\displaystyle {\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}}
corrigé
Nous pouvons commencer par transformer le second membre en utilisant l'identité remarquable :
A
3
−
B
3
=
(
A
−
B
)
(
A
2
+
A
B
+
B
2
)
{\displaystyle A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{2})}
En posant :
{
A
=
a
+
b
B
=
a
−
b
{\displaystyle {\begin{cases}A={\sqrt {a+b}}\\B={\sqrt {a-b}}\end{cases}}}
On a alors :
(
a
+
b
)
3
−
(
a
−
b
)
3
=
(
a
+
b
)
3
−
(
a
−
b
)
3
=
(
a
+
b
−
a
−
b
)
(
a
+
b
+
a
+
b
a
−
b
+
a
−
b
)
=
(
a
+
b
−
a
−
b
)
(
2
a
+
a
2
−
b
2
)
{\displaystyle {\begin{aligned}{\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}&=\left({\sqrt {a+b}}\right)^{3}-\left({\sqrt {a-b}}\right)^{3}\\&=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(a+b+{\sqrt {a+b}}{\sqrt {a-b}}+a-b\right)\\&=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(2a+{\sqrt {a^{2}-b^{2}}}\right)\\\end{aligned}}}
En remplaçant dans :
2
(
2
a
+
a
2
−
b
2
)
a
−
a
2
−
b
2
=
(
a
+
b
)
3
−
(
a
−
b
)
3
{\displaystyle {\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}}
on obtient :
2
(
2
a
+
a
2
−
b
2
)
a
−
a
2
−
b
2
=
(
a
+
b
−
a
−
b
)
(
2
a
+
a
2
−
b
2
)
{\displaystyle {\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(2a+{\sqrt {a^{2}-b^{2}}}\right)}
qui se simplifie par
(
2
a
+
a
2
−
b
2
)
{\displaystyle \left(2a+{\sqrt {a^{2}-b^{2}}}\right)}
2
a
−
a
2
−
b
2
=
a
+
b
−
a
−
b
{\displaystyle {\sqrt {2}}{\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {a+b}}-{\sqrt {a-b}}}
qui s'écrit aussi :
a
−
a
2
−
b
2
=
a
+
b
2
−
a
−
b
2
{\displaystyle {\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {\frac {a+b}{2}}}-{\sqrt {\frac {a-b}{2}}}}
et l'on remarque que l'on obtient directement cette relation en utilisant l'identité remarquable (vu en cours) :
A
−
B
=
A
+
A
2
−
B
2
−
A
−
A
2
−
B
2
{\displaystyle {\sqrt {A-{\sqrt {B}}}}={\sqrt {\frac {A+{\sqrt {A^{2}-B}}}{2}}}-{\sqrt {\frac {A-{\sqrt {A^{2}-B}}}{2}}}}
en posant cette fois :
{
A
=
a
B
=
a
2
−
b
2
{\displaystyle {\begin{cases}A=a\\B=a^{2}-b^{2}\end{cases}}}
![{\displaystyle {\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3934f8c1ae01f0b3b00b871ad5ec02de41e61dac)
![{\displaystyle {\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7620d32f72402854e439bf36e18f047786162bbe)
![{\displaystyle {\begin{aligned}{\frac {1}{2{\sqrt {2}}+{\sqrt {3}}+{\sqrt {10}}+{\sqrt {15}}}}&={\frac {(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})}{\left[(2{\sqrt {2}}+{\sqrt {10}})+({\sqrt {3}}+{\sqrt {15}})\right]\left[(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})\right]}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(2{\sqrt {2}}+{\sqrt {10}})^{2}-({\sqrt {3}}+{\sqrt {15}})^{2}}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(8+8{\sqrt {5}}+10)-(3+6{\sqrt {5}}+15)}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{2{\sqrt {5}}}}\\&={\frac {(2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}){\sqrt {5}}}{2{\sqrt {5}}{\sqrt {5}}}}\\&={\frac {2{\sqrt {10}}+5{\sqrt {2}}-{\sqrt {15}}-5{\sqrt {3}}}{10}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20127820ecce4162b63d5e8aca889100e18a8cb7)
![{\displaystyle a=1\qquad \qquad b={\sqrt[{3}]{2}}\qquad \qquad c={\sqrt[{3}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/228bcca9137bc27aa40e5926025a8a437a7f85cf)
![{\displaystyle {\begin{aligned}{\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left(1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {1+{\sqrt[{3}]{4}}+{\sqrt[{3}]{16}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-{\sqrt[{3}]{8}}}{1+2+4-3{\sqrt[{3}]{1\times 2\times 4}}}}\\&={\frac {1+{\sqrt[{3}]{4}}+2{\sqrt[{3}]{2}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-2}{7-3{\sqrt[{3}]{8}}}}\\&={\frac {{\sqrt[{3}]{2}}-1}{7-3\times 2}}\\&={\sqrt[{3}]{2}}-1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd0c7fbedf817bb066f47aa88c056b4791af4482)
![{\displaystyle {\begin{aligned}{\frac {1}{{\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}}-{\sqrt {7}}}}&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{\left[({\sqrt {2}}-{\sqrt {3}})+({\sqrt {5}}-{\sqrt {7}})\right]\left[({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})\right]}}\\&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{({\sqrt {2}}-{\sqrt {3}})^{2}-({\sqrt {5}}-{\sqrt {7}})^{2}}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{(5-2{\sqrt {6}})-(12-2{\sqrt {35}})}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{2{\sqrt {35}}-7-2{\sqrt {6}}}}\\&={\frac {\left[{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}\right]\left[2{\sqrt {35}}+7+2{\sqrt {6}}\right]}{\left[2{\sqrt {35}}-(7+2{\sqrt {6}})\right]\left[2{\sqrt {35}}+(7+2{\sqrt {6}})\right]}}\\&={\frac {2{\sqrt {70}}+7{\sqrt {2}}+4{\sqrt {3}}-2{\sqrt {105}}-7{\sqrt {3}}-6{\sqrt {2}}-10{\sqrt {7}}-7{\sqrt {5}}-2{\sqrt {30}}+14{\sqrt {5}}+7{\sqrt {7}}+2{\sqrt {42}}}{140-(7+2{\sqrt {6}})^{2}}}\\&={\frac {{\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}}{67-28{\sqrt {6}}}}\\&={\frac {\left({\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}\right)\left(67+28{\sqrt {6}}\right)}{\left(67-28{\sqrt {6}}\right)\left(67+28{\sqrt {6}}\right)}}\\&={\frac {-185{\sqrt {2}}-145{\sqrt {3}}+133{\sqrt {5}}+135{\sqrt {7}}+62{\sqrt {30}}+50{\sqrt {42}}-34{\sqrt {70}}-22{\sqrt {105}}}{-215}}\\&={\frac {185{\sqrt {2}}+145{\sqrt {3}}-133{\sqrt {5}}-135{\sqrt {7}}-62{\sqrt {30}}-50{\sqrt {42}}+34{\sqrt {70}}+22{\sqrt {105}}}{215}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c420653f5bd5848893a51924496330b64dd8a16)
![{\displaystyle a={\sqrt[{3}]{2}}\qquad \qquad b={\sqrt[{3}]{3}}\qquad \qquad c={\sqrt[{3}]{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdab6df74e7ecf2d26e7c9095648e2cf56006002)
![{\displaystyle {\begin{aligned}{\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left({\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{2+3+5-3{\sqrt[{3}]{2\times 3\times 5}}}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{10-3{\sqrt[{3}]{30}}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5181512d1bf543bbbb6397c7acf767750cc75294)
![{\displaystyle a=10\qquad \qquad b=3{\sqrt[{3}]{30}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00e9326a730bd608cdabd73af77f22b4e4cba3de)
![{\displaystyle {\begin{aligned}\qquad &={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)(a^{2}+ab+b^{2})}{\left(10-3{\sqrt[{3}]{30}}\right)(a^{2}+ab+b^{2})}}\\&={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)\left(100+30{\sqrt[{3}]{3}}0+9{\sqrt[{3}]{900}}\right))}{10^{3}-\left(3{\sqrt[{3}]{30}}\right)^{3}}}\\&={\frac {100{\sqrt[{3}]{4}}+60{\sqrt[{3}]{15}}+18{\sqrt[{3}]{450}}+100{\sqrt[{3}]{9}}+90{\sqrt[{3}]{10}}+27{\sqrt[{3}]{300}}+100{\sqrt[{3}]{25}}+150{\sqrt[{3}]{6}}+45{\sqrt[{3}]{180}}-100{\sqrt[{3}]{6}}-30{\sqrt[{3}]{180}}-54{\sqrt[{3}]{25}}-100{\sqrt[{3}]{10}}-30{\sqrt[{3}]{300}}-90{\sqrt[{3}]{9}}-100{\sqrt[{3}]{15}}-30{\sqrt[{3}]{450}}-135{\sqrt[{3}]{4}}}{1000-810}}\\&={\frac {-35{\sqrt[{3}]{4}}+50{\sqrt[{3}]{6}}+10{\sqrt[{3}]{9}}-10{\sqrt[{3}]{10}}-40{\sqrt[{3}]{15}}+46{\sqrt[{3}]{25}}+15{\sqrt[{3}]{180}}-3{\sqrt[{3}]{300}}-12{\sqrt[{3}]{450}}}{190}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6c5e5a1f6f2c90f4d217510ce2c7c7b835f8ef9)
![{\displaystyle {\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}={\sqrt[{3}]{\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65270c0c500fb6048ffbde4793e369e498751e68)
![{\displaystyle {\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}={\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f46353b78a6dedeedef967bc11581604d0c0f861)
![{\displaystyle {\sqrt[{3}]{38+17{\sqrt {5}}}}={\sqrt {9+4{\sqrt {5}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dd81d00374fe8fc3135d2788fdccf6f8b8ca7af)
![{\displaystyle {\begin{aligned}\left({\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}\right)^{3}&={\frac {1^{3}+3{\sqrt {3}}+3({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{(2{\sqrt[{3}]{2}})^{3}}}\\&={\frac {1+3{\sqrt {3}}+9+3{\sqrt {3}}}{16}}\\&={\frac {10+6{\sqrt {3}}}{16}}\\&={\frac {5+3{\sqrt {3}}}{8}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b43366efbd81ed3876ce74c6f6607123c54f037)
![{\displaystyle {\begin{aligned}\left({\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}\right)^{3}&={\frac {2^{3}+3\times 2^{2}{\sqrt {3}}+3\times 2({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{20+12{\sqrt {3}}}}\\&={\frac {8+12{\sqrt {3}}+18+3{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {26+15{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {(26+15{\sqrt {3}})(20-12{\sqrt {3}})}{(20+12{\sqrt {3}})(20-12{\sqrt {3}})}}\\&={\frac {520-312{\sqrt {3}}+300{\sqrt {3}}-540}{400-432}}\\&={\frac {-20-12{\sqrt {3}}}{-32}}\\&={\frac {5+3{\sqrt {3}}}{8}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2250c872a9a9025ef36abba46280d4506e103de6)
![{\displaystyle {\begin{aligned}\left({\sqrt[{3}]{38+17{\sqrt {5}}}}\right)^{6}&=\left(38+17{\sqrt {5}}\right)^{2}\\&=38^{2}+2\times 38\times 17{\sqrt {5}}+\left(17{\sqrt {5}}\right)^{2}\\&=2889+1292{\sqrt {5}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df4338ddf9ac4337cd6c079949eb3eee6d27f444)
![{\displaystyle {\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23aea5ebabeb0f249c7f25e15f256ab2024d0e49)
![{\displaystyle {\begin{aligned}{\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}&={\frac {\sqrt {\left({\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}\right)^{2}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}-2{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}+{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {2{\sqrt[{4}]{8}}-2{\sqrt {\left({\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}\right)\left({\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {8}}-\left({\sqrt {2}}-1\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {2{\sqrt {2}}-{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baf1a94da7d6634f4e76d55d07aa5ca41e9aa69a)
![{\displaystyle {\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(x-2)(x+2)}}}{(x-1)(x+2)+(x+1){\sqrt {(x-2)(x+2)}}}}\\&={\frac {(x+1){\sqrt {x-2}}{\sqrt {x-2}}+(x-1){\sqrt {x-2}}{\sqrt {x+2}}}{(x-1){\sqrt {x+2}}{\sqrt {x+2}}+(x+1){\sqrt {x-2}}{\sqrt {x+2}}}}\\&={\frac {{\sqrt {x-2}}\left[(x+1){\sqrt {x-2}}+(x-1){\sqrt {x+2}}\right]}{{\sqrt {x+2}}\left[(x-1){\sqrt {x+2}}+(x+1){\sqrt {x-2}}\right]}}\\&={\frac {\sqrt {x-2}}{\sqrt {x+2}}}\\&={\sqrt {\frac {x-2}{x+2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cda9396d51d6ed3901e8527619fe64c91e4b34af)
![{\displaystyle {\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(2-x)(-2-x)}}}{(x-1)(x+2)+(x+1){\sqrt {(2-x)(-2-x)}}}}\\&={\frac {-(x+1){\sqrt {2-x}}{\sqrt {2-x}}+(x-1){\sqrt {2-x}}{\sqrt {-2-x}}}{-(x-1){\sqrt {-2-x}}{\sqrt {-2-x}}+(x+1){\sqrt {2-x}}{\sqrt {-2-x}}}}\\&={\frac {{\sqrt {2-x}}\left[-(x+1){\sqrt {2-x}}+(x-1){\sqrt {-2-x}}\right]}{{\sqrt {-2-x}}\left[-(x-1){\sqrt {-2-x}}+(x+1){\sqrt {2-x}}\right]}}\\&=-{\frac {\sqrt {2-x}}{\sqrt {-2-x}}}\\&=-{\sqrt {\frac {2-x}{-2-x}}}\\&=-{\sqrt {\frac {x-2}{x+2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0653e857e1aaa4dc6ed33aed9c1f579ea6a2ad8)
