Question

\(\sqrt{2x-1}-\sqrt{\frac{x+3}{2x-1}}\le\sqrt{4x-3}-\sqrt{\frac{x+3}{4x-3}}\)

Answer 1

ĐKXĐ: \(x\ge\frac{3}{4}\)

\(\sqrt{4x-3}-\sqrt{2x-1}+\sqrt{\frac{x+3}{2x-1}}-\sqrt{\frac{x+3}{4x-3}}\ge0\)

\(\Rightarrow\sqrt{4x-3}-\sqrt{2x-1}+\sqrt{x+3}\left(\frac{\sqrt{4x-3}-\sqrt{2x-1}}{\sqrt{4x-3}.\sqrt{2x-1}}\right)\ge0\)

\(\Rightarrow\left(\sqrt{4x-3}-\sqrt{2x-1}\right)\left(1+\frac{\sqrt{x+3}}{\sqrt{4x-3}\sqrt{2x-1}}\right)\ge0\)

\(\Rightarrow\sqrt{4x-3}-\sqrt{2x-1}\ge0\) (do \(1+\frac{\sqrt{x+3}}{\sqrt{4x-3}\sqrt{2x-1}}>0\))

\(\Rightarrow\sqrt{4x-3}\ge\sqrt{2x-1}\)

\(\Rightarrow4x-3\ge2x-1\)

\(\Rightarrow x\ge1\)