GG
H24
5 tháng 5 lúc 9:52

ĐKXĐ: \(x\ge0;x\ne4\)

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

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{x+4}{\sqrt{x}+2}\)

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

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

Bình luận (0)