Question

Chứng minh:

(\(\dfrac{99x+1}{5x^2-5}\) + \(\dfrac{1}{5+5x}\) + \(\dfrac{20}{1-x}\)) : \(\dfrac{4}{x^3y-xy}\) = -5xy

Answer 1

\(\left(\dfrac{99x+1}{5x^2-5}+\dfrac{1}{5+5x}+\dfrac{20}{1-x}\right):\dfrac{4}{x^3y-xy}\)

\(=\left(\dfrac{99x+1}{5\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{5\left(x-1\right)\left(x+1\right)}-\dfrac{100\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\right):\dfrac{4}{xy\left(x^2-1\right)}\)

\(=\dfrac{99x+1+x-1-100x-100}{5\left(x-1\right)\left(x+1\right)}:\dfrac{4}{xy\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-100xy}{20}=-5xy=VP\)( đpcm )