Question

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

Answer 1

Lời giải:
ĐKXĐ: $x\neq -1; x\neq 0; x\neq 2$

\(Q=1+\left[\frac{x+1}{(x+1)(x^2-x+1)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2(x-2)}{x(x^2-x+1)}\)

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

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

\(=\frac{x(x+1)-(2x^2-2x+2)}{(x+1)(x-2)}=\frac{-x^2+3x-2}{(x+1)(x-2)}=\frac{(1-x)(x-2)}{(x+1)(x-2)}=\frac{1-x}{1+x}\)