Question

Rút gọn biểu thức sau:

  

Answer 1

ĐK: \(\left\{{}\begin{matrix}x\ne1\\x\ne0\\x\ne2\end{matrix}\right.\)

Khi đó:

\(B=\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right):\left(\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\right)\\ =\dfrac{1}{x\left(x-1\right)}:\left(\dfrac{x^2-1-x^2+4}{\left(x-2\right)\left(x-1\right)}\right)\\ =\dfrac{1}{x\left(x-1\right)}.\dfrac{\left(x-2\right)\left(x-1\right)}{3}\\ =\dfrac{x-2}{3x}\)

Answer 2

\(B=\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right):\left(\dfrac{x+1}{x-2}-\dfrac{x+2}{x-1}\right)\)

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

\(=\dfrac{1}{x\left(x-1\right)}\cdot\dfrac{\left(x-2\right)\left(x-1\right)}{x^2-1-x^2+4}\)

\(=\dfrac{\left(x-2\right)}{3x}\)