Question

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

 

a) CM: P= \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b) Tìm các giá trị của x để 2P= 2\(\sqrt{x}\) + 5

Answer 1

a, Với x >= 0 x khác 1 

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

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

Vậy ta có đpcm 

b, \(2P=2\sqrt{x}+5\Rightarrow P=\sqrt{x}+\dfrac{5}{2}\)

\(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\Rightarrow x=\dfrac{1}{4}\)