ĐKXĐ: \(x\ge9\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x-9}=\sqrt{x-1}+\sqrt{x-4}\)
\(\Leftrightarrow2x-9+2\sqrt{x^2-9x}=2x-5+2\sqrt{x^2-4x+3}\)
\(\Leftrightarrow\sqrt{x^2-9x}=2+\sqrt{x^2-4x+3}\)
Do \(x\ge9>0\Rightarrow x^2-4x>x^2-9x\Rightarrow x^2-4x+3>x^2-9x\)
\(\Rightarrow\sqrt{x^2-4x+3}+2>\sqrt{x^2-9x}\)
Pt vô nghiệm