a)\(A=\frac{x+1}{x^2-2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\left(\frac{1}{x\left(x-1\right)}+\frac{1}{x-1}\right)\)
\(=\frac{x+1}{\left(x-1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{x+1}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{1+x}\)
\(=\frac{x}{x-1}\)
b)Có:\(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=2\end{array}\right.\)
Thay x=2 vào A ta được:
\(A=\frac{2}{2-1}=2\)
