Question

[Azaki]

A = \(\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\) (x >0; x ≠ 4)

a, Rút gọn

b, Tìm x để A > \(\frac{1}{2}\)

Answer 1

[Akai Haruma]

Lời giải:

a) \(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)

b)

\(A>\frac{1}{2}\Leftrightarrow \frac{2}{\sqrt{x}+2}>\frac{1}{2}\Leftrightarrow 4> \sqrt{x}+2\Leftrightarrow 4> x\geq 0\)

Kết hợp với ĐKXĐ suy ra $4>x>0$