Question

[Thảo Nguyễn]

A=(\(\frac{x-2}{x+2\sqrt{x}}\)+\(\frac{1}{\sqrt{x}+2}\)).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

a) rút gọn 

tìm x để 2A = 2\(\sqrt{5}\)+5

Answer 1

[Trần Thị Thảo Ngọc]

a, \(A=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}=\left(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x-1+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

Answer 2

[Trần Thị Thảo Ngọc]

Để \(2A=2\sqrt{5}+5\) thì : \(2.\frac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{5}+5\)

\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{5}+5\)

\(\Leftrightarrow2\sqrt{x}+2=\left(2\sqrt{5}\right)x+5\sqrt{x}\)

\(\Leftrightarrow\left(2\sqrt{5}\right)x+3\sqrt{x}=2\)\(\Leftrightarrow\left(2\sqrt{5}+3\right)\sqrt{x}=2\Leftrightarrow\sqrt{x}=\frac{2}{2\sqrt{5}+3}=\frac{4\sqrt{5}-6}{11}\Leftrightarrow x=\left(\frac{4\sqrt{5}-6}{11}\right)^2\left(tm\right)\)

Vậy \(x=\left(\frac{4\sqrt{5}-6}{11}\right)^2\)thì \(2A=2\sqrt{5}+5\)