Question

[nood]

Cho biểu thức C=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\dfrac{x-4}{\sqrt{4x}}\) với x>0 và x khác 4 

a) Rút gọn C 

b) Tìm x để C>3

Answer 1

[乇尺尺のﾚ]

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

b) C>3

\(\Rightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\)