Question

Cho biểu thức :

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

với x > 0 , x ≠ 1

a. Rút gọn A

b. Tìm x để A = 1/3

Answer 1

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

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

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

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b) \(\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-\sqrt{x}=3\)

\(\Leftrightarrow2\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{9}{4}\)