Question

Cho biểu thức:

A = \((\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2})\frac{\sqrt{x}-2}{2}\)

a, Rút gọn A

b, Tìm x nguyên để 2A có giá trị nguyên

Answer 1

ĐKXĐ: \(x\ge0;x\ne4\)

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

\(=\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(2A=\frac{2\sqrt{x}+2}{\sqrt{x}+2}=\frac{2\sqrt{x}+4-2}{\sqrt{x}+2}=2-\frac{2}{\sqrt{x}+2}\)

Để 2A nguyên \(\Rightarrow\frac{2}{\sqrt{x}+2}\) nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(2\right)\)

Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=2\Rightarrow x=0\)