Question

Cho \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)

Tìm giá trị của P khi \(x=\dfrac{2}{2-\sqrt{3}}-2\sqrt{3}\)

Answer 1

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{x-1}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}=\dfrac{x+2\sqrt{x}}{x-1}.\dfrac{x\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}=\dfrac{x}{\sqrt{x}-1}\) ( x > 0 ; x # 1 )

Ta có : \(x=\dfrac{2}{2-\sqrt{3}}-2\sqrt{3}=\dfrac{2-4\sqrt{3}+6}{2-\sqrt{3}}=\dfrac{8-4\sqrt{3}}{2-\sqrt{3}}=\dfrac{4\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=4\left(TMĐKXĐ\right)\)Khi đó : \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{4}{2-1}=4\)