Question

Rút gọn :

A= ( \(\dfrac{x+2}{32}+\dfrac{2}{x+1}-3\) ).\(\dfrac{x+1}{2-4x}-\dfrac{3x-x^2+1}{3x}\)

Answer 1

\(=\dfrac{x^2+3x+2+64-96\left(x+1\right)}{32\left(x+1\right)}\cdot\dfrac{x+1}{-2\left(x-2\right)}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{x^2+3x+66-96x-96}{32}\cdot\dfrac{-1}{2\left(x-2\right)}+\dfrac{x^2-3x-1}{3x}\)

\(=-\dfrac{x^2-93x-30}{64\left(x-2\right)}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{-3x\left(x^2-93x-30\right)}{192x\left(x-2\right)}+\dfrac{64\left(x-2\right)\left(x^2-3x-1\right)}{192x\left(x-2\right)}\)

\(=\dfrac{-3x^3+279x^2+90x+64\left(x^3-3x^2-x-2x^2+6x+2\right)}{192x\left(x-2\right)}\)

\(=\dfrac{-3x^3+279x^2+90x+64x^3-320x^2+320x+128}{192x\left(x-2\right)}\)

\(=\dfrac{61x^3-41x^2+410x+128}{192x\left(x-2\right)}\)