Lời giải:
ĐKXĐ: \(x\neq \pm 1; x\neq 0; x\neq \pm \sqrt{2}\)
a)
\(A=\left(\frac{(x-1)(x^2+x+1)}{x-1}+x\right)\left(\frac{(x+1)(x^2-x+1)}{x+1}-x\right).\frac{x^2-2}{x(x^2-1)^2}\)
\(=(x^2+x+1+x)(x^2-x+1-x).\frac{x^2-2}{x[(x-1)(x+1)]^2}\)
\(=(x^2+2x+1)(x^2-2x+1).\frac{x^2-2}{x(x-1)^2(x+1)^2}\)
\(=(x+1)^2(x-1)^2.\frac{x^2-2}{x(x-1)^2(x+1)^2}\)
\(=\frac{x^2-2}{x}\)
b) Khi \(x=2\Rightarrow A=\frac{2^2-2}{2}=1\)
c)
\(A=-1\Leftrightarrow \frac{x^2-2}{x}=-1\)
\(\Leftrightarrow x^2-2=-x\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\)
\(\Rightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Kết hợp với ĐKXĐ ban đầu ta suy ra $x=-2$
