Question

cho biểu thức 

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

a. rút gọn P

b.tính gtri biểu thức P khi x=\(\dfrac{1}{\sqrt{2}}\)

Answer 1

\(a,=\dfrac{1}{\sqrt{x}+1}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

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

\(=\dfrac{-x-1}{x-1}\)

Vậy\(P=\dfrac{-x-1}{x-1}\)

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

\(P=\dfrac{-\left(\dfrac{1}{\sqrt{2}}\right)-1}{\dfrac{1}{\sqrt{2}}-1}=\dfrac{-\sqrt{2}}{2}\)

Vậy \(P=\dfrac{-\sqrt{2}}{2}\) khi \(x=\dfrac{1}{\sqrt{2}}\)