Question

Giải hộ mik vs ạ

 

  

Answer 1

\(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ =\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}\\ =\dfrac{x+2\sqrt{x}}{x-4}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để \(A>1\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}>1\Leftrightarrow\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0\Leftrightarrow2>0\left(LD\right)\)

\(\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow x>4\left(tm\right)\)

Vậy \(x>4\) thì \(A>1\).