Question

\(B=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right);\dfrac{4}{x-4}\) (với x < 0; x khác 0)

Answer 1

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\cdot\dfrac{x-4}{4}=\dfrac{2\sqrt{x}}{4}=\dfrac{1}{2}\sqrt{x}\)

Answer 2

Với x < 0 ; x ≠ 0 ta có:

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

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\cdot\dfrac{x-4}{4}\)

\(=\dfrac{2\sqrt{x}}{4}=\dfrac{\sqrt{x}}{2}\)

Vậy \(B=\dfrac{\sqrt{x}}{2}\).