En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : ÉchauffementCalcul littéral/Exercices/Échauffement », n'a pu être restituée correctement ci-dessus.
Ne pas confondre addition et multiplication
modifier
x
×
x
=
{\displaystyle x\times x=}
Solution
x
×
x
=
x
2
{\displaystyle x\times x=x^{2}}
x
+
x
=
…
{\displaystyle x+x=\ldots }
Solution
x
+
x
=
2
x
{\displaystyle x+x=2x}
2
x
+
3
x
=
…
{\displaystyle 2x+3x=\ldots }
Solution
2
x
+
3
x
=
5
x
{\displaystyle 2x+3x=5x}
−
2
x
+
3
x
=
…
{\displaystyle -2x+3x=\ldots }
Solution
−
2
x
+
3
x
=
1
x
=
x
{\displaystyle -2x+3x=1x=x}
2
x
−
3
x
=
…
{\displaystyle 2x-3x=\ldots }
Solution
2
x
−
3
x
=
−
1
x
=
−
x
{\displaystyle 2x-3x=-1x=-x}
−
2
x
−
3
x
=
…
{\displaystyle -2x-3x=\ldots }
Solution
−
2
x
−
3
x
=
−
5
x
{\displaystyle -2x-3x=-5x}
2
×
x
×
3
=
…
{\displaystyle 2\times x\times 3=\ldots }
Solution
2
×
x
×
3
=
2
×
3
×
x
=
6
×
x
=
6
x
{\displaystyle 2\times x\times 3=2\times 3\times x=6\times x=6x}
2
×
x
×
(
−
3
)
=
…
{\displaystyle 2\times x\times (-3)=\ldots }
Solution
2
×
x
×
(
−
3
)
=
2
×
(
−
3
)
×
x
=
−
6
×
x
=
−
6
x
{\displaystyle 2\times x\times (-3)=2\times (-3)\times x=-6\times x=-6x}
2
×
x
×
3
x
=
…
{\displaystyle 2\times x\times 3x=\ldots }
Solution
2
×
x
×
3
x
=
2
×
3
×
x
×
x
=
6
×
x
2
=
6
x
2
{\displaystyle 2\times x\times 3x=2\times 3\times x\times x=6\times x^{2}=6x^{2}}
2
×
x
×
(
−
3
x
)
=
…
{\displaystyle 2\times x\times (-3x)=\ldots }
Solution
2
×
x
×
(
−
3
x
)
=
2
×
(
−
3
)
×
x
×
x
=
−
6
×
x
2
=
−
6
x
2
{\displaystyle 2\times x\times (-3x)=2\times (-3)\times x\times x=-6\times x^{2}=-6x^{2}}
(
2
x
)
2
=
…
{\displaystyle (2x)^{2}=\ldots }
Solution
(
2
x
)
2
=
2
2
×
x
2
=
4
x
2
{\displaystyle (2x)^{2}=2^{2}\times x^{2}=4x^{2}}
(
−
x
)
2
=
…
{\displaystyle (-x)^{2}=\ldots }
Solution
Il faut utiliser la règle :
(
a
×
b
)
2
=
a
2
×
b
2
{\displaystyle (a\times b)^{2}=a^{2}\times b^{2}}
(
−
x
)
2
=
(
−
1
×
x
)
2
=
(
−
1
)
2
×
x
2
=
1
×
x
2
=
x
2
{\displaystyle (-x)^{2}=(-1\times x)^{2}=(-1)^{2}\times x^{2}=1\times x^{2}=x^{2}}
3
(
5
x
)
2
=
…
{\displaystyle 3(5x)^{2}=\ldots }
Solution
3
(
5
x
)
2
=
3
×
5
2
×
x
2
=
3
×
25
×
x
2
=
75
x
2
{\displaystyle 3(5x)^{2}=3\times 5^{2}\times x^{2}=3\times 25\times x^{2}=75x^{2}}
−
(
3
x
)
2
=
…
{\displaystyle -(3x)^{2}=\ldots }
Solution
−
(
3
x
)
2
=
−
3
2
×
x
2
=
−
9
x
2
{\displaystyle -(3x)^{2}=-3^{2}\times x^{2}=-9x^{2}}
modifier
3
x
+
5
x
2
−
5
x
=
…
{\displaystyle 3x+5x^{2}-5x=\ldots }
Solution
3
x
+
5
x
2
−
5
x
=
5
x
2
+
3
x
−
5
x
=
5
x
2
+
(
3
−
5
)
x
=
5
x
2
−
2
x
{\displaystyle 3x+5x^{2}-5x=5x^{2}+3x-5x=5x^{2}+(3-5)x=5x^{2}-2x}
−
3
x
2
+
5
x
2
−
5
x
=
…
{\displaystyle -3x^{2}+5x^{2}-5x=\ldots }
Solution
−
3
x
2
+
5
x
2
−
5
x
=
(
−
3
+
5
)
x
2
−
5
x
=
2
x
2
−
5
x
{\displaystyle -3x^{2}+5x^{2}-5x=(-3+5)x^{2}-5x=2x^{2}-5x}
3
x
−
1
−
5
x
2
−
5
x
+
2
=
…
{\displaystyle 3x-1-5x^{2}-5x+2=\ldots }
Solution
3
x
−
1
−
5
x
2
−
5
x
+
2
=
−
5
x
2
+
(
3
−
5
)
x
+
2
−
1
=
−
5
x
2
−
2
x
+
1
{\displaystyle 3x-1-5x^{2}-5x+2=-5x^{2}+(3-5)x+2-1=-5x^{2}-2x+1}
1
2
x
+
5
x
2
−
5
x
=
…
{\displaystyle {\frac {1}{2}}x+5x^{2}-5x=\ldots }
Solution
1
2
x
+
5
x
2
−
5
x
=
5
x
2
+
(
1
2
−
5
)
x
=
5
x
2
+
(
1
2
−
10
2
)
x
=
5
x
2
−
9
2
x
{\displaystyle {\frac {1}{2}}x+5x^{2}-5x=5x^{2}+({\frac {1}{2}}-5)x=5x^{2}+({\frac {1}{2}}-{\frac {10}{2}})x=5x^{2}-{\frac {9}{2}}x}
1
2
x
2
+
5
x
2
−
5
x
=
…
{\displaystyle {\frac {1}{2}}x^{2}+5x^{2}-5x=\ldots }
Solution
1
2
x
2
+
5
x
2
−
5
x
=
(
1
2
+
5
)
x
2
−
5
x
=
(
1
2
+
10
2
)
x
2
−
5
x
=
11
2
x
2
−
5
x
{\displaystyle {\frac {1}{2}}x^{2}+5x^{2}-5x=({\frac {1}{2}}+5)x^{2}-5x=({\frac {1}{2}}+{\frac {10}{2}})x^{2}-5x={\frac {11}{2}}x^{2}-5x}
2
3
x
+
5
x
2
−
5
x
−
2
5
x
2
=
…
{\displaystyle {\frac {2}{3}}x+5x^{2}-5x-{\frac {2}{5}}x^{2}=\ldots }
Solution
2
3
x
+
5
x
2
−
5
x
−
2
5
x
2
=
(
5
−
2
5
)
x
2
+
(
2
3
−
5
)
x
=
(
25
5
−
2
5
)
x
2
+
(
2
3
−
15
3
)
x
=
23
5
x
2
−
13
3
x
{\displaystyle {\frac {2}{3}}x+5x^{2}-5x-{\frac {2}{5}}x^{2}=(5-{\frac {2}{5}})x^{2}+({\frac {2}{3}}-5)x=({\frac {25}{5}}-{\frac {2}{5}})x^{2}+({\frac {2}{3}}-{\frac {15}{3}})x={\frac {23}{5}}x^{2}-{\frac {13}{3}}x}
