Exercice 00-07
Résoudre dans
R
\mathbb{R}
R
:
2
x
x
+
1
=
2
x
−
1
x
\frac{2x}{x+1}=\frac{2x-1}{x}
x
+
1
2
x
=
x
2
x
−
1
x
2
−
x
−
12
=
0
x^2-x-12=0
x
2
−
x
−
12
=
0
2
x
2
+
4
x
+
1
=
0
2x^2+4x+1=0
2
x
2
+
4
x
+
1
=
0
x
4
−
3
x
2
+
2
=
0
x^4-3x^2+2=0
x
4
−
3
x
2
+
2
=
0
3
∣
x
−
4
∣
=
10
3\,|x-4|=10
3
∣
x
−
4∣
=
10
2
x
4
−
x
=
3
4
−
x
\frac{2x}{\sqrt{4-x}}=3\sqrt{4-x}
4
−
x
2
x
=
3
4
−
x
(
x
+
1
)
(
x
+
6
)
=
1
(x+1)(x+6)=1
(
x
+
1
)
(
x
+
6
)
=
1
∣
∣
x
−
1
∣
−
3
∣
=
1
\bigl||x-1|-3\bigr|=1
∣
x
−
1∣
−
3
=
1
Indications
Forum
Solution
S
=
{
1
}
S=\{1\}
S
=
{
1
}
S
=
{
−
3
,
4
}
S=\{-3,4\}
S
=
{
−
3
,
4
}
S
=
{
−
1
±
2
2
}
S=\{-1\pm \frac{\sqrt{2}}{2}\}
S
=
{
−
1
±
2
2
}
S
=
{
±
1
,
±
2
}
S=\{\pm 1,\pm\sqrt{2}\}
S
=
{
±
1
,
±
2
}
S
=
{
22
3
,
2
3
}
S=\{\frac{22}{3},\frac{2}{3}\}
S
=
{
3
22
,
3
2
}
S
=
{
12
5
}
S=\{\frac{12}{5}\}
S
=
{
5
12
}
S
=
{
−
7
±
29
2
}
S=\{\frac{-7\pm \sqrt{29}}{2}\}
S
=
{
2
−
7
±
29
}
S
=
{
−
3
,
−
1
,
3
,
5
}
S=\{-3,-1,3,5\}
S
=
{
−
3
,
−
1
,
3
,
5
}