Question

Cho E = \(\dfrac{\sqrt{x}}{x+2\sqrt{x}}\): \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

a) Rút gọn biểu thức E 

b) Tìm x khi E = \(\dfrac{2}{5}\)

Answer 1

a: ĐKXĐ: x>0

\(E=\dfrac{\sqrt{x}}{x+2\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

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

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

b: E=2/5

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)

=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)

=>\(2x-3\sqrt{x}+4=0\)

=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)

=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)

=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)

Vậy: \(x\in\varnothing\)