En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : QuotientIdentités remarquables/Exercices/Quotient », n'a pu être restituée correctement ci-dessus.
Simplifier les fractions :
a)
x
2
−
1
x
2
−
2
x
+
1
{\displaystyle {\frac {x^{2}-1}{x^{2}-2x+1}}}
Solution
x
2
−
1
x
2
−
2
x
+
1
=
(
x
+
1
)
(
x
−
1
)
(
x
−
1
)
2
=
x
+
1
x
−
1
{\displaystyle {\frac {x^{2}-1}{x^{2}-2x+1}}={\frac {(x+1)(x-1)}{(x-1)^{2}}}={\frac {x+1}{x-1}}}
b)
x
2
−
1
x
2
+
2
x
+
1
{\displaystyle {\frac {x^{2}-1}{x^{2}+2x+1}}}
Solution
x
2
−
1
x
2
+
2
x
+
1
=
(
x
+
1
)
(
x
−
1
)
(
x
+
1
)
2
=
x
−
1
x
+
1
{\displaystyle {\frac {x^{2}-1}{x^{2}+2x+1}}={\frac {(x+1)(x-1)}{(x+1)^{2}}}={\frac {x-1}{x+1}}}
c)
x
2
−
4
x
+
4
x
2
−
4
{\displaystyle {\frac {x^{2}-4x+4}{x^{2}-4}}}
Solution
x
2
−
4
x
+
4
x
2
−
4
=
x
2
−
2
×
2
x
+
2
2
x
2
−
2
2
=
(
x
−
2
)
2
(
x
+
2
)
(
x
−
2
)
=
x
−
2
x
+
2
{\displaystyle {\frac {x^{2}-4x+4}{x^{2}-4}}={\frac {x^{2}-2\times 2x+2^{2}}{x^{2}-2^{2}}}={\frac {(x-2)^{2}}{(x+2)(x-2)}}={\frac {x-2}{x+2}}}
d)
(
2
x
−
1
)
2
−
(
x
+
1
)
2
x
2
−
4
x
+
4
{\displaystyle {\frac {(2x-1)^{2}-(x+1)^{2}}{x^{2}-4x+4}}}
Solution
(
2
x
−
1
)
2
−
(
x
+
1
)
2
x
2
−
4
x
+
4
=
(
2
x
−
1
+
x
+
1
)
(
2
x
−
1
−
x
−
1
)
x
2
−
2
×
2
x
+
2
2
=
3
x
(
x
−
2
)
(
x
−
2
)
2
=
3
x
x
−
2
{\displaystyle {\frac {(2x-1)^{2}-(x+1)^{2}}{x^{2}-4x+4}}={\frac {(2x-1+x+1)(2x-1-x-1)}{x^{2}-2\times 2x+2^{2}}}={\frac {3x(x-2)}{(x-2)^{2}}}={\frac {3x}{x-2}}}
e)
x
2
+
6
x
+
9
−
(
2
x
+
5
)
2
(
2
x
+
3
)
2
−
(
x
+
1
)
2
{\displaystyle {\frac {x^{2}+6x+9-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}}
Solution
x
2
+
6
x
+
9
−
(
2
x
+
5
)
2
(
2
x
+
3
)
2
−
(
x
+
1
)
2
=
(
x
+
3
)
2
−
(
2
x
+
5
)
2
(
2
x
+
3
)
2
−
(
x
+
1
)
2
=
(
x
+
3
+
2
x
+
5
)
(
x
+
3
−
2
x
−
5
)
(
2
x
+
3
+
x
+
1
)
(
2
x
+
3
−
x
−
1
)
=
(
3
x
+
8
)
(
−
x
−
2
)
(
3
x
+
4
)
(
x
+
2
)
=
−
(
3
x
+
8
)
(
x
+
2
)
(
3
x
+
4
)
(
x
+
2
)
=
−
3
x
+
8
3
x
+
4
{\displaystyle {\begin{aligned}{\frac {x^{2}+6x+9-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}&={\frac {(x+3)^{2}-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}\\&={\frac {(x+3+2x+5)(x+3-2x-5)}{(2x+3+x+1)(2x+3-x-1)}}\\&={\frac {(3x+8)(-x-2)}{(3x+4)(x+2)}}\\&={\frac {-(3x+8)(x+2)}{(3x+4)(x+2)}}\\&=-{\frac {3x+8}{3x+4}}\end{aligned}}}
f)
(
3
a
−
b
)
2
−
(
4
a
+
b
)
2
3
a
2
b
+
6
a
b
2
{\displaystyle {\frac {(3a-b)^{2}-(4a+b)^{2}}{3a^{2}b+6ab^{2}}}}
Solution
(
3
a
−
b
)
2
−
(
4
a
+
b
)
2
3
a
2
b
+
6
a
b
2
=
(
3
a
−
b
+
4
a
+
b
)
(
3
a
−
b
−
4
a
−
b
)
3
a
b
(
a
+
2
b
)
=
7
a
(
−
a
−
2
b
)
3
a
b
(
a
+
2
b
)
=
−
7
a
(
a
+
2
b
)
3
a
b
(
a
+
2
b
)
=
−
7
3
b
{\displaystyle {\begin{aligned}{\frac {(3a-b)^{2}-(4a+b)^{2}}{3a^{2}b+6ab^{2}}}&={\frac {(3a-b+4a+b)(3a-b-4a-b)}{3ab(a+2b)}}\\&={\frac {7a(-a-2b)}{3ab(a+2b)}}\\&={\frac {-7a(a+2b)}{3ab(a+2b)}}\\&={\frac {-7}{3b}}\end{aligned}}}
Effectuer les calculs :
a)
2
x
2
−
1
+
3
x
+
1
{\displaystyle {\frac {2}{x^{2}-1}}+{\frac {3}{x+1}}}
Solution
2
x
2
−
1
+
3
x
+
1
=
2
x
2
−
1
+
3
(
x
−
1
)
(
x
+
1
)
(
x
−
1
)
=
2
+
3
(
x
−
1
)
x
2
−
1
=
3
x
−
1
x
2
−
1
{\displaystyle {\frac {2}{x^{2}-1}}+{\frac {3}{x+1}}={\frac {2}{x^{2}-1}}+{\frac {3(x-1)}{(x+1)(x-1)}}={\frac {2+3(x-1)}{x^{2}-1}}={\frac {3x-1}{x^{2}-1}}}
b)
3
x
2
−
1
−
9
2
(
x
−
1
)
(
x
−
2
)
{\displaystyle {\frac {3}{x^{2}-1}}-{\frac {9}{2(x-1)(x-2)}}}
Solution
3
x
2
−
1
−
9
2
(
x
−
1
)
(
x
−
2
)
=
3
(
x
+
1
)
(
x
−
1
)
−
9
2
(
x
−
1
)
(
x
−
2
)
=
2
×
3
(
x
−
2
)
2
(
x
+
1
)
(
x
−
1
)
(
x
−
2
)
−
9
(
x
+
1
)
2
(
x
−
1
)
(
x
+
1
)
(
x
−
2
)
=
6
(
x
−
2
)
−
9
(
x
+
1
)
2
(
x
+
1
)
(
x
−
1
)
(
x
−
2
)
=
−
3
x
−
21
2
(
x
+
1
)
(
x
−
1
)
(
x
−
2
)
=
−
3
(
x
+
7
)
2
(
x
+
1
)
(
x
−
1
)
(
x
−
2
)
{\displaystyle {\begin{aligned}{\frac {3}{x^{2}-1}}-{\frac {9}{2(x-1)(x-2)}}&={\frac {3}{(x+1)(x-1)}}-{\frac {9}{2(x-1)(x-2)}}\\&={\frac {2\times 3(x-2)}{2(x+1)(x-1)(x-2)}}-{\frac {9(x+1)}{2(x-1)(x+1)(x-2)}}\\&={\frac {6(x-2)-9(x+1)}{2(x+1)(x-1)(x-2)}}\\&={\frac {-3x-21}{2(x+1)(x-1)(x-2)}}\\&={\frac {-3(x+7)}{2(x+1)(x-1)(x-2)}}\end{aligned}}}
c)
2
x
+
1
2
x
2
−
8
x
+
8
−
x
+
3
x
2
−
4
{\displaystyle {\frac {2x+1}{2x^{2}-8x+8}}-{\frac {x+3}{x^{2}-4}}}
Solution
2
x
+
1
2
x
2
−
8
x
+
8
−
x
+
3
x
2
−
4
=
2
x
+
1
2
(
x
2
−
4
x
+
2
2
)
−
x
+
3
x
2
−
2
2
=
2
x
+
1
2
(
x
−
2
)
2
−
x
+
3
(
x
+
2
)
(
x
−
2
)
=
(
2
x
+
1
)
(
x
+
2
)
2
(
x
−
2
)
2
(
x
+
2
)
−
2
(
x
−
2
)
(
x
+
3
)
2
(
x
+
2
)
(
x
−
2
)
2
=
(
2
x
+
1
)
(
x
+
2
)
−
2
(
x
−
2
)
(
x
+
3
)
2
(
x
+
2
)
(
x
−
2
)
2
=
2
x
2
+
4
x
+
x
+
2
−
2
x
2
−
6
x
+
4
x
+
12
2
(
x
+
2
)
(
x
−
2
)
2
=
3
x
+
14
2
(
x
+
2
)
(
x
−
2
)
2
{\displaystyle {\begin{aligned}{\frac {2x+1}{2x^{2}-8x+8}}-{\frac {x+3}{x^{2}-4}}&={\frac {2x+1}{2(x^{2}-4x+2^{2})}}-{\frac {x+3}{x^{2}-2^{2}}}\\&={\frac {2x+1}{2(x-2)^{2}}}-{\frac {x+3}{(x+2)(x-2)}}\\&={\frac {(2x+1)(x+2)}{2(x-2)^{2}(x+2)}}-{\frac {2(x-2)(x+3)}{2(x+2)(x-2)^{2}}}\\&={\frac {(2x+1)(x+2)-2(x-2)(x+3)}{2(x+2)(x-2)^{2}}}\\&={\frac {2x^{2}+4x+x+2-2x^{2}-6x+4x+12}{2(x+2)(x-2)^{2}}}\\&={\frac {3x+14}{2(x+2)(x-2)^{2}}}\end{aligned}}}
d)
1
1
−
x
+
1
1
+
x
−
4
x
2
1
−
x
4
{\displaystyle {\frac {1}{1-x}}+{\frac {1}{1+x}}-{\frac {4x^{2}}{1-x^{4}}}}
Solution
1
1
−
x
+
1
1
+
x
−
4
x
2
1
−
x
4
=
1
+
x
(
1
−
x
)
(
1
+
x
)
+
1
−
x
(
1
+
x
)
(
1
−
x
)
−
4
x
2
1
−
x
4
=
1
+
x
1
−
x
2
+
1
−
x
1
−
x
2
−
4
x
2
1
−
x
4
=
1
+
x
+
1
−
x
1
−
x
2
−
4
x
2
1
−
x
4
=
2
1
−
x
2
−
4
x
2
1
−
x
4
=
2
(
1
+
x
2
)
(
1
−
x
2
)
(
1
+
x
2
)
−
4
x
2
1
−
x
4
=
2
+
2
x
2
1
−
x
4
−
4
x
2
1
−
x
4
=
2
−
2
x
2
1
−
x
4
=
2
(
1
−
x
2
)
(
1
+
x
2
)
(
1
−
x
2
)
=
2
1
+
x
2
{\displaystyle {\begin{aligned}{\frac {1}{1-x}}+{\frac {1}{1+x}}-{\frac {4x^{2}}{1-x^{4}}}&={\frac {1+x}{(1-x)(1+x)}}+{\frac {1-x}{(1+x)(1-x)}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {1+x}{1-x^{2}}}+{\frac {1-x}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {1+x+1-x}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2(1+x^{2})}{(1-x^{2})(1+x^{2})}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2+2x^{2}}{1-x^{4}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2-2x^{2}}{1-x^{4}}}\\&={\frac {2(1-x^{2})}{(1+x^{2})(1-x^{2})}}\\&={\frac {2}{1+x^{2}}}\end{aligned}}}
e)
x
4
x
4
−
1
−
x
2
x
2
−
1
+
1
x
−
1
−
1
x
+
1
{\displaystyle {\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {1}{x-1}}-{\frac {1}{x+1}}}
Solution
x
4
x
4
−
1
−
x
2
x
2
−
1
+
1
x
−
1
−
1
x
+
1
=
x
4
x
4
−
1
−
x
2
x
2
−
1
+
x
+
1
x
2
−
1
−
x
−
1
x
2
−
1
=
x
4
x
4
−
1
−
x
2
x
2
−
1
+
2
x
2
−
1
=
x
4
x
4
−
1
+
2
−
x
2
x
2
−
1
=
x
4
x
4
−
1
+
(
2
−
x
2
)
(
x
2
+
1
)
(
x
2
−
1
)
(
x
2
+
1
)
=
x
4
x
4
−
1
+
2
x
2
+
2
−
x
4
−
x
2
x
4
−
1
=
x
2
+
2
x
4
−
1
{\displaystyle {\begin{aligned}{\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {1}{x-1}}-{\frac {1}{x+1}}&={\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {x+1}{x^{2}-1}}-{\frac {x-1}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {2}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {2-x^{2}}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {(2-x^{2})(x^{2}+1)}{(x^{2}-1)(x^{2}+1)}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {2x^{2}+2-x^{4}-x^{2}}{x^{4}-1}}\\&={\frac {x^{2}+2}{x^{4}-1}}\end{aligned}}}
