\(\left(x+1\right)^2\left(x-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x+1=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1=-1\\x=0+1=1\end{cases}}\)
Vậy: \(x\in\left\{1;-1\right\}\)
(x+1)2.(x-1)2=0
=> x+1=0 hoac x-1=0
=> x=0-1 hoac x=0+1
=> x=-1 hoac x=1
=>\(\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\)
+ (x-1)2=0
=>(x-1)2=02
=>x-1=0
=>x=0+1
x=1
+(x+1)2=0
=>x+1=0 (k thỏa mãn)
=>x=1
vay x=1
k mình nha