Question

Cho biểu thức: với x ≥ 0 và x ≠ 4

A= \(\left(\sqrt{x}-\frac{x+2\sqrt{x}+3}{\sqrt{x}+3}\right).\left(1+\frac{5}{\sqrt{x}-2}\right)\)
a) Rút gọn A
b) Tìm x để A>1

Answer 1

a/ \(A=\left(\frac{x+3\sqrt{x}-x-2\sqrt{x}-3}{\sqrt{x}+3}\right)\left(\frac{\sqrt{x}-2+5}{\sqrt{x}-2}\right)\)

\(=\frac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\frac{\sqrt{x}+3}{\sqrt{x}-2}=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)

b/ A>1 <=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}>1\Leftrightarrow\frac{\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\frac{-1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)