ĐKXĐ: x∉{0;1;-1}
a: \(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x^2-1+x+2-x^2}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x-1}\)
b: Để A<0 thì \(\frac{x^2}{x-1}<0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được: \(\begin{cases}x<1\\ x\notin\left\lbrace0;-1\right\rbrace\end{cases}\)
c: \(A=\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1}=\left(x+1\right)+\frac{1}{x-1}=x-1+\frac{1}{x-1}+2\)
=>\(A\ge2\cdot\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+2=2+2=4\forall x\) >1
Dấu '=' xảy ra khi x-1=1
=>x=2
