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Fonction exponentielle
Sommes nous à l'aise avec les formules usuelles des exponentielles - Exercice 2
5 min
15
Question 1
Montrer que, pour tout réel
x
x
x
, on a :
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
1
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2}=1
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
1
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
5
x
×
e
2
x
e
(
−
x
+
1
)
×
3
×
e
5
x
−
1
×
e
(
−
5
2
x
+
1
)
×
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =\frac{e^{5x} \times e^{2x} }{e^{\left(-x+1\right)\times 3} \times e^{5x-1} } \times e^{\left(-\frac{5}{2} x+1\right)\times 2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
(
−
x
+
1
)
×
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
e
(
−
2
5
x
+
1
)
×
2
équivaut successivement à :
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
5
x
×
e
2
x
e
−
3
x
+
3
×
e
5
x
−
1
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =\frac{e^{5x} \times e^{2x} }{e^{-3x+3} \times e^{5x-1} } \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
−
3
x
+
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
5
x
+
2
x
e
−
3
x
+
3
+
5
x
−
1
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =\frac{e^{5x+2x} }{e^{-3x+3+5x-1} } \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
−
3
x
+
3
+
5
x
−
1
e
5
x
+
2
x
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
7
x
e
2
x
+
2
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =\frac{e^{7x} }{e^{2x+2} } \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
2
x
+
2
e
7
x
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
7
x
−
(
2
x
+
2
)
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =e^{7x-\left(2x+2\right)} \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
7
x
−
(
2
x
+
2
)
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
7
x
−
2
x
−
2
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =e^{7x-2x-2} \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
7
x
−
2
x
−
2
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
5
x
−
2
×
e
−
5
x
+
2
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =e^{5x-2} \times e^{-5x+2}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
5
x
−
2
×
e
−
5
x
+
2
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
5
x
−
2
+
(
−
5
x
+
2
)
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =e^{5x-2+\left(-5x+2\right)}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
5
x
−
2
+
(
−
5
x
+
2
)
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
e
0
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =e^{0}
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
e
0
Ainsi :
e
5
x
×
e
2
x
(
e
−
x
+
1
)
3
×
e
5
x
−
1
×
(
e
−
5
2
x
+
1
)
2
=
1
\frac{e^{5x} \times e^{2x} }{\left(e^{-x+1} \right)^{3} \times e^{5x-1} } \times \left(e^{-\frac{5}{2} x+1} \right)^{2} =1
(
e
−
x
+
1
)
3
×
e
5
x
−
1
e
5
x
×
e
2
x
×
(
e
−
2
5
x
+
1
)
2
=
1