ĐKXĐ: \(x\ge\frac{3}{2}\)
\(PT\Leftrightarrow\frac{2x-3-x}{\sqrt{2x-3}+\sqrt{x}}-2\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{2x-3}+\sqrt{x}}-2\right)=0\Leftrightarrow x=3\)
hoặc \(\frac{1}{\sqrt{2x-3}+\sqrt{x}}-2=0\Leftrightarrow2\sqrt{2x-3}+2\sqrt{x}=1\). Ta có:\(x\ge\frac{3}{2}\)
\(\Rightarrow\sqrt{2x-3}+2\sqrt{x}\ge0+2\sqrt{\frac{3}{2}}=\sqrt{6}>1\Rightarrow\)vô nghiệm
Vậy PT nghiệm duy nhất x = 3
\(\sqrt{2x-3}-\sqrt{x}=2x-6.\)
\(\Rightarrow...\)