Question

Cho biểu thức A= \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)( với x≥0, x≠4)

a. Rút gọn A

b. Tìm x để A>0

Answer 1

a/ \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{-1}{\sqrt{x}-2}\)

b/ \(A>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0\)

Ta thấy: - 1 < 0 nên để A > 0 thì:

\(\sqrt{x}-2< 0\)\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

kết hợp với đkxđ: => \(0\le x< 4\)