a: Ta có: \(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\)
\(=\frac{\left(x-1\right)^2}{x^2-2x+1+3x}+\frac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\)
\(=\frac{\left(x-1\right)^2}{x^2+x+1}+\frac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\)
\(=\frac{\left(x-1\right)^3+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}=1\)
Ta có: \(A=\left\lbrack\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right\rbrack:\frac{x^2+x}{x^3+x}\)
\(=\frac{1\cdot\left(x^3+x\right)}{x^2+x}=\frac{x^2+1}{x+1}\)
b: A>-1
=>A+1>0
=>\(\frac{x^2+1+x+1}{x+1}>0\)
=>\(\frac{x^2+x+2}{x+1}>0\)
=>x+1>0
=>x>-1
Kết hợp ĐKXĐ, ta được: x>-1 và x<>1
