Question

Cho biểu thức: \(Q=\frac{\sqrt{x}+1}{2\sqrt{x}-2}-\frac{\sqrt{x}-1}{2\sqrt{x}+2}-\frac{2}{\sqrt{x}-1}\)

a) Rút gọn biểu thức b) Tìm x để \(Q\)>1

Answer 1

ĐKXĐ: \(x\ge0;x\ne1\)

\(Q=\frac{\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{4\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

Để \(Q>1\Rightarrow\frac{-2}{x-1}>1\Rightarrow\frac{x+1}{x-1}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)

Vậy \(0\le x< 1\)