Ta có:[tex]\sqrt{4x+5}-\sqrt{3x+1}=\frac{4x+5-3x-1}{\sqrt{4x+5}+\sqrt{3x+1}}=\frac{x+4}{\sqrt{4x+5}+\sqrt{3x+1}}=\frac{x+4}{9}\Leftrightarrow (x+4)(\frac{1}{\sqrt{4x+5}+\sqrt{3x+1}}-\frac{1}{9})=0\Leftrightarrow x=-4(loại)hoặc\sqrt{4x+5}+\sqrt{3x+1}=9\Leftrightarrow 4x+5+3x+1+2\sqrt{12x^2+19x+5}=9\Leftrightarrow 2\sqrt{12x^2+19x+5}=3-7x\Leftrightarrow 48x^2+76x+20=49x^2-42x+9\Leftrightarrow x^2-118x-11=0\Leftrightarrow x=59\pm 6\sqrt{97}[/tex]