Question

Answer 1

Bài 4: 

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

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

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

\(=\dfrac{-2}{x-2}\)

b: Thay x=1 vào B, ta được:

\(B=\dfrac{-2}{1-2}=\dfrac{-2}{-1}=2\)

c: Để B<0 thì x-2<0

hay x<2

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< 2\\x\notin\left\{-2;0\right\}\end{matrix}\right.\)