Question

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

Tính P khi \(x=\dfrac{1}{\sqrt{2}}\)

Answer 1

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

Thay : \(x=\dfrac{1}{\sqrt{2}}\left(TM\right)\) vào P , ta có :

\(\left(\dfrac{1}{\sqrt{2}}+1\right):\left(1-\dfrac{1}{\sqrt{2}}\right)=\dfrac{\sqrt{2}+1}{\sqrt{2}}.\dfrac{\sqrt{2}}{\sqrt{2}-1}=\left(\sqrt{2}+1\right)^2\)