Question

Rút gọn biểu thức:A=\(\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{4}{x-4}\)

Answer 1

đk : x >= 0 ; x khác 4 

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

Answer 2

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

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

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

Answer 3

ĐKXĐ: x khác 4; x ≥ 0

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

Answer 4

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