Question

Rút gọn Biểu thức A= ( √x/√x+3 + √x/√x-3) / 2√x/x-9

Answer 1

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{2\sqrt{x}}{x-9}\) (ĐKXĐ: \(x>0;x\ne9\))

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\left[\dfrac{x-3\sqrt{x}+x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{2\sqrt{x}}\)

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

\(=\sqrt{x}\)

Answer 2

đk : x khác 9 ; x > 0 

\(=\dfrac{x-3\sqrt{x}+x+3\sqrt{x}}{x-9}:\dfrac{2\sqrt{x}}{x-9}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)