Question

 Cho biểu thức :\(A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\) ,với x > 0 

a) Rút gọn biểu thức A            b) Tìm giá trị tất cả của x để A ≥ \(\dfrac{1}{3\sqrt{x}}\)

Answer 1

`a.`\(A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\) \(\left(x>0\right)\)

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

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

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

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

`b.`\(A\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)\le3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}+2\le3\)

\(\Leftrightarrow\sqrt{x}\le1\)

\(\Leftrightarrow x\le1\)

Vậy \(S=\left\{x|0< x\le1\right\}\)