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Calcul numérique : les puissances, le calcul fractionnaire et les racines carrées
Savoir effectuer des calculs numériques utilisant des puissances - Exercice 1
20 min
40
Simplifier au maximum les expressions suivantes :
Question 1
A
=
1
0
2
×
1
0
5
A=10^{2} \times 10^{5}
A
=
1
0
2
×
1
0
5
Correction
1
0
a
×
1
0
b
=
1
0
a
+
b
10^{a} \times 10^{b} =10^{a+b}
1
0
a
×
1
0
b
=
1
0
a
+
b
1
0
a
1
0
b
=
1
0
a
−
b
\frac{10^{a} }{10^{b} } =10^{a-b}
1
0
b
1
0
a
=
1
0
a
−
b
(
1
0
a
)
b
=
1
0
a
×
b
\left(10^{a} \right)^{b} =10^{a\times b}
(
1
0
a
)
b
=
1
0
a
×
b
1
0
−
a
=
1
1
0
a
10^{-a} =\frac{1}{10^{a} }
1
0
−
a
=
1
0
a
1
x
a
×
x
b
=
x
a
+
b
x^{a}\times x^{b} =x^{a+b}
x
a
×
x
b
=
x
a
+
b
(
x
y
)
a
=
x
a
y
a
\left(\frac{x}{y} \right)^{a} =\frac{x^{a} }{y^{a} }
(
y
x
)
a
=
y
a
x
a
(
x
y
)
a
=
x
a
×
y
a
\left(xy\right)^{a} =x^{a} \times y^{a}
(
x
y
)
a
=
x
a
×
y
a
A
=
1
0
2
×
1
0
5
A=10^{2} \times 10^{5}
A
=
1
0
2
×
1
0
5
équivaut successivement à :
A
=
1
0
2
+
5
A=10^{2+5}
A
=
1
0
2
+
5
A
=
1
0
7
A=10^{7}
A
=
1
0
7
Question 2
B
=
1
0
4
×
1
0
−
3
×
1
0
6
B=10^{4} \times 10^{-3}\times 10^{6}
B
=
1
0
4
×
1
0
−
3
×
1
0
6
Correction
1
0
a
×
1
0
b
=
1
0
a
+
b
10^{a} \times 10^{b} =10^{a+b}
1
0
a
×
1
0
b
=
1
0
a
+
b
1
0
a
1
0
b
=
1
0
a
−
b
\frac{10^{a} }{10^{b} } =10^{a-b}
1
0
b
1
0
a
=
1
0
a
−
b
(
1
0
a
)
b
=
1
0
a
×
b
\left(10^{a} \right)^{b} =10^{a\times b}
(
1
0
a
)
b
=
1
0
a
×
b
1
0
−
a
=
1
1
0
a
10^{-a} =\frac{1}{10^{a} }
1
0
−
a
=
1
0
a
1
(
x
y
)
a
=
x
a
y
a
\left(\frac{x}{y} \right)^{a} =\frac{x^{a} }{y^{a} }
(
y
x
)
a
=
y
a
x
a
(
x
y
)
a
=
x
a
×
y
a
\left(xy\right)^{a} =x^{a} \times y^{a}
(
x
y
)
a
=
x
a
×
y
a
B
=
1
0
4
×
1
0
−
3
×
1
0
6
B=10^{4} \times 10^{-3}\times 10^{6}
B
=
1
0
4
×
1
0
−
3
×
1
0
6
équivaut successivement à :
B
=
1
0
4
+
(
−
3
)
+
6
B=10^{4+\left(-3\right)+6}
B
=
1
0
4
+
(
−
3
)
+
6
B
=
1
0
4
−
3
+
6
B=10^{4-3+6}
B
=
1
0
4
−
3
+
6
B
=
1
0
7
B=10^{7}
B
=
1
0
7
Question 3
C
=
1
0
5
×
1
0
8
1
0
4
C=\frac{10^{5} \times 10^{8} }{10^{4} }
C
=
1
0
4
1
0
5
×
1
0
8
Correction
1
0
a
×
1
0
b
=
1
0
a
+
b
10^{a} \times 10^{b} =10^{a+b}
1
0
a
×
1
0
b
=
1
0
a
+
b
1
0
a
1
0
b
=
1
0
a
−
b
\frac{10^{a} }{10^{b} } =10^{a-b}
1
0
b
1
0
a
=
1
0
a
−
b
(
1
0
a
)
b
=
1
0
a
×
b
\left(10^{a} \right)^{b} =10^{a\times b}
(
1
0
a
)
b
=
1
0
a
×
b
1
0
−
a
=
1
1
0
a
10^{-a} =\frac{1}{10^{a} }
1
0
−
a
=
1
0
a
1
(
x
y
)
a
=
x
a
y
a
\left(\frac{x}{y} \right)^{a} =\frac{x^{a} }{y^{a} }
(
y
x
)
a
=
y
a
x
a
(
x
y
)
a
=
x
a
×
y
a
\left(xy\right)^{a} =x^{a} \times y^{a}
(
x
y
)
a
=
x
a
×
y
a
C
=
1
0
5
×
1
0
8
1
0
4
C=\frac{10^{5} \times 10^{8} }{10^{4} }
C
=
1
0
4
1
0
5
×
1
0
8
équivaut successivement à :
C
=
1
0
5
+
8
1
0
4
C=\frac{10^{5+8} }{10^{4} }
C
=
1
0
4
1
0
5
+
8
C
=
1
0
13
1
0
4
C=\frac{10^{13} }{10^{4} }
C
=
1
0
4
1
0
13
C
=
1
0
13
−
4
C=10^{13-4}
C
=
1
0
13
−
4
C
=
1
0
9
C=10^{9}
C
=
1
0
9
Question 4
D
=
(
1
0
2
)
3
×
1
0
−
1
(
1
0
2
×
1
0
5
)
2
D=\frac{\left(10^{2} \right)^{3} \times 10^{-1} }{\left(10^{2} \times 10^{5} \right)^{2} }
D
=
(
1
0
2
×
1
0
5
)
2
(
1
0
2
)
3
×
1
0
−
1
Correction
1
0
a
×
1
0
b
=
1
0
a
+
b
10^{a} \times 10^{b} =10^{a+b}
1
0
a
×
1
0
b
=
1
0
a
+
b
1
0
a
1
0
b
=
1
0
a
−
b
\frac{10^{a} }{10^{b} } =10^{a-b}
1
0
b
1
0
a
=
1
0
a
−
b
(
1
0
a
)
b
=
1
0
a
×
b
\left(10^{a} \right)^{b} =10^{a\times b}
(
1
0
a
)
b
=
1
0
a
×
b
1
0
−
a
=
1
1
0
a
10^{-a} =\frac{1}{10^{a} }
1
0
−
a
=
1
0
a
1
D
=
(
1
0
2
)
3
×
1
0
−
1
(
1
0
2
×
1
0
5
)
2
D=\frac{\left(10^{2} \right)^{3} \times 10^{-1} }{\left(10^{2} \times 10^{5} \right)^{2} }
D
=
(
1
0
2
×
1
0
5
)
2
(
1
0
2
)
3
×
1
0
−
1
équivaut successivement à :
D
=
1
0
2
×
3
×
1
0
−
1
(
1
0
2
+
5
)
2
D=\frac{10^{2\times 3} \times 10^{-1} }{\left(10^{2+5} \right)^{2} }
D
=
(
1
0
2
+
5
)
2
1
0
2
×
3
×
1
0
−
1
D
=
1
0
6
×
1
0
−
1
(
1
0
7
)
2
D=\frac{10^{6} \times 10^{-1} }{\left(10^{7} \right)^{2} }
D
=
(
1
0
7
)
2
1
0
6
×
1
0
−
1
D
=
1
0
6
+
(
−
1
)
1
0
7
×
2
D=\frac{10^{6+\left(-1\right)} }{10^{7\times 2} }
D
=
1
0
7
×
2
1
0
6
+
(
−
1
)
D
=
1
0
5
1
0
14
D=\frac{10^{5} }{10^{14} }
D
=
1
0
14
1
0
5
D
=
1
0
5
−
14
D=10^{5-14}
D
=
1
0
5
−
14
D
=
1
0
−
9
D=10^{-9}
D
=
1
0
−
9
Question 5
E
=
3
6
×
3
−
4
×
3
2
3
3
×
3
E=\frac{3^6\times 3^{-4}\times 3^2}{3^3\times 3}
E
=
3
3
×
3
3
6
×
3
−
4
×
3
2
Correction
Soit
x
x
x
un réel non nul.
x
a
×
x
b
=
x
a
+
b
x^{a} \times x^{b} =x^{a+b}
x
a
×
x
b
=
x
a
+
b
x
a
x
b
=
x
a
−
b
\frac{x^{a} }{x^{b} } =x^{a-b}
x
b
x
a
=
x
a
−
b
x
0
=
1
x^{0} =1
x
0
=
1
E
=
3
6
×
3
−
4
×
3
2
3
3
×
3
E=\frac{3^6\times 3^{-4}\times 3^2}{3^3\times 3}
E
=
3
3
×
3
3
6
×
3
−
4
×
3
2
équivaut successivement à :
E
=
3
6
×
3
−
4
×
3
2
3
3
×
3
1
E=\frac{3^6\times 3^{-4}\times 3^2}{3^3\times 3^1}
E
=
3
3
×
3
1
3
6
×
3
−
4
×
3
2
E
=
3
6
+
(
−
4
)
+
2
3
3
+
1
E=\frac{3^{6+\left(-4\right)+2}}{3^{3+1}}
E
=
3
3
+
1
3
6
+
(
−
4
)
+
2
E
=
3
4
3
4
E=\frac{3^4}{3^4}
E
=
3
4
3
4
E
=
3
4
−
4
E=3^{4-4}
E
=
3
4
−
4
E
=
3
0
E=3^0
E
=
3
0
Ainsi :
E
=
1
E=1
E
=
1
Question 6
F
=
(
4
2
)
3
(
4
−
1
)
6
×
(
4
−
3
)
4
(
4
5
)
2
F=\frac{{\left(4^2\right)}^3}{{\left(4^{-1}\right)}^6}\times \frac{{\left(4^{-3}\right)}^4}{{\left(4^5\right)}^2}
F
=
(
4
−
1
)
6
(
4
2
)
3
×
(
4
5
)
2
(
4
−
3
)
4
Correction
Soit
x
x
x
un réel non nul.
x
a
×
x
b
=
x
a
+
b
x^{a} \times x^{b} =x^{a+b}
x
a
×
x
b
=
x
a
+
b
x
a
x
b
=
x
a
−
b
\frac{x^{a} }{x^{b} } =x^{a-b}
x
b
x
a
=
x
a
−
b
(
x
a
)
b
=
x
a
×
b
\left(x^{a} \right)^{b} =x^{a\times b}
(
x
a
)
b
=
x
a
×
b
F
=
(
4
2
)
3
(
4
−
1
)
6
×
(
4
−
3
)
4
(
4
5
)
2
F=\frac{{\left(4^2\right)}^3}{{\left(4^{-1}\right)}^6}\times \frac{{\left(4^{-3}\right)}^4}{{\left(4^5\right)}^2}
F
=
(
4
−
1
)
6
(
4
2
)
3
×
(
4
5
)
2
(
4
−
3
)
4
équivaut successivement à :
F
=
4
2
×
3
4
(
−
1
)
×
6
×
4
(
−
3
)
×
4
4
5
×
2
F=\frac{4^{2\times 3}}{4^{\left(-1\right)\times 6}}\times \frac{4^{\left(-3\right)\times 4}}{4^{5\times 2}}
F
=
4
(
−
1
)
×
6
4
2
×
3
×
4
5
×
2
4
(
−
3
)
×
4
F
=
4
6
4
−
6
×
4
−
12
4
10
F=\frac{4^6}{4^{-6}}\times \frac{4^{-12}}{4^{10}}
F
=
4
−
6
4
6
×
4
10
4
−
12
F
=
4
6
−
(
−
6
)
×
4
−
12
−
10
F=4^{6-\left(-6\right)}\times 4^{-12-10}
F
=
4
6
−
(
−
6
)
×
4
−
12
−
10
F
=
4
6
+
6
×
4
−
12
−
10
F=4^{6+6}\times 4^{-12-10}
F
=
4
6
+
6
×
4
−
12
−
10
F
=
4
12
×
4
−
22
F=4^{12}\times 4^{-22}
F
=
4
12
×
4
−
22
F
=
4
12
+
(
−
22
)
F=4^{12+\left(-22\right)}
F
=
4
12
+
(
−
22
)
Ainsi :
F
=
4
−
10
F=4^{-10}
F
=
4
−
10
Question 7
G
=
(
(
10
−
4
)
3
)
2
G={\left({\left({10}^{-4}\right)}^3\right)}^2
G
=
(
(
10
−
4
)
3
)
2
Correction
Soit
x
x
x
un réel non nul.
(
x
a
)
b
=
x
a
×
b
\left(x^{a} \right)^{b} =x^{a\times b}
(
x
a
)
b
=
x
a
×
b
G
=
(
(
10
−
4
)
3
)
2
G={\left({\left({10}^{-4}\right)}^3\right)}^2
G
=
(
(
10
−
4
)
3
)
2
G
=
(
10
(
−
4
)
×
3
)
2
G={\left({10}^{\left(-4\right)\times 3}\right)}^2
G
=
(
10
(
−
4
)
×
3
)
2
G
=
(
10
−
12
)
2
G={\left({10}^{-12}\right)}^2
G
=
(
10
−
12
)
2
G
=
10
(
−
12
)
×
2
G={10}^{\left(-12\right)\times 2}
G
=
10
(
−
12
)
×
2
Ainsi :
G
=
10
−
24
G={10}^{-24}
G
=
10
−
24