ĐKXĐ: \(x\notin\left\{1;0;-1\right\}\)
a: \(\dfrac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{x^2-2x+1+3x}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\)
\(=\dfrac{\left(x-1\right)^3+2x^2-4x-1+\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}=1\)
\(A=\left[\dfrac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right]:\dfrac{x^2+x}{x^3+x}\)
\(=1:\dfrac{x\left(x+1\right)}{x\left(x^2+1\right)}=\dfrac{x^2+1}{x+1}\)
b: A>-1
=>A+1>0
=>\(\dfrac{x^2+1+x+1}{x+1}>0\)
=>\(\dfrac{x^2+x+2}{x+1}>0\)
mà \(x^2+x+2=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}>0\forall x\)
nên x+1>0
=>x>-1
