Question

Answer 1

a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)

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

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

\(=\dfrac{-4x}{\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)

\(=\dfrac{-4x^2}{-x+3}=\dfrac{4x^2}{x-3}\)

b: Để B<0 thì x-3<0

hay x<3

Kết hợp ĐKXĐ, ta có: \(\left\{{}\begin{matrix}x< 3\\x\notin\left\{0;2;-2\right\}\end{matrix}\right.\)