Ta có: |x + 1| = |x - 1|
<=> x + 1 = x - 1
=> x + x =1 - 1
=> 2x = 0
=> x =0
Ta có: 4 x |x| = x2 + 4
=> |x| = (x2 + 4) : 4
=> |x| = x2 : 4 + 1
=> x = 1/4x2 + 1
=> x - 1/4x2 = 1
=> x.(1 - 1/4x) = 1
=> x . 2/3 = 1
=
a) \(\left|x+1\right|=\left|x-1\right|\Leftrightarrow\hept{\begin{cases}x+1=x-1\\x+1=-\left(x-1\right)=-x+1\end{cases}}\)
+)x+1=x-1
=>x+1-(x-1)=0
=>x+1-x+1=0=>(x-x)+2=0=>0+2=0=>o=-2(vô lí)
+)x+1=-x+1
=>x+1-(-x+1)=0
=>x+1+x-1=0=>2x=0=>x=0
Vậy x=0 thì thỏa mãn PT