Question

cho biểu thức . rút gọn

A=\(\left(\dfrac{x^3-1}{x-1}+x\right)\left(\dfrac{x^3+1}{x+1}-x\right):\dfrac{x\left(1-x\right)^2}{x^2-2}\)

Answer 1

Lời giải:

ĐK: \(x\neq 0;x\neq \pm 1; x\neq \pm \sqrt{2}\)

Ta có:
\(A=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x(1-x)^2}{x^2-2}\)

\(=\left(\frac{(x-1)(x^2+x+1)}{x-1}+x\right)\left(\frac{(x+1)(x^2-x+1)}{x+1}-x\right).\frac{x^2-2}{x(x-1)^2}\)

\(=(x^2+x+1+x)(x^2-x+1-x).\frac{x^2-2}{x(x-1)^2}\)

\(=(x^2+2x+1)(x^2-2x+1).\frac{x^2-2}{x(x-1)^2}\)

\(=(x+1)^2(x-1)^2.\frac{x^2-2}{x(x-1)^2}=(x+1)^2.\frac{x^2-2}{x}\)