Lời giải:
ĐKXĐ: $-1\leq x\leq 1$
PT \(\Rightarrow \left\{\begin{matrix}
x-1\geq 0\\
1-x^2=(x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\geq 1\\
(x-1)^2+(x^2-1)=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\geq 1\\
(x-1)(x-1+x+1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\ge 1\\ 2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy ..........