Question

Cho B = (\(\dfrac{4x}{x+2}-\dfrac{x^3-8}{x^3+8}\cdot\dfrac{4x^2-8x+16}{x^2-4}\)):\(\dfrac{16}{x+2}\cdot\dfrac{x^2+3x+2}{x^2+x+1}\)

1) Rút gọn B

2) Tìm x để B = 0.

Answer 1

1: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{16}{x+2}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)

\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)

\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)

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

2: Để B=0 thì -x-1=0

hay x=-1(nhận)