Question

[Linh Nhi]

Cho biểu thức P=\(\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}\) (Với x>0, x\(\ne\)1)

a.Rút gọn P

b.Tính gá trị biểu thức P khi x=\(\dfrac{1}{\sqrt{2}}\)

Answer 1

[Như]

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

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

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

b/ thay x = \(\dfrac{1}{\sqrt{2}}\) vào P:

\(P=\dfrac{\dfrac{1}{\sqrt{2}}+1}{1-\dfrac{1}{\sqrt{2}}}=3+2\sqrt{2}\)