Question

\(\sqrt{10x+1}\) + \(\sqrt{3x-5}\) = \(\sqrt{9x+4}\) + \(\sqrt{2x-2}\)

Answer 1

ĐK: \(x\ge\dfrac{5}{3}\)

\(pt\Leftrightarrow\left(\sqrt{10x+1}-\sqrt{9x+4}\right)+\left(\sqrt{3x-5}-\sqrt{2x-2}\right)=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)

Dễ thấy \(\dfrac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\dfrac{1}{\sqrt{3x-5}+\sqrt{2x-2}}>0\)

\(pt\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)