ĐK \(x\ge\frac{2}{3}\)
Đặt \(\hept{\begin{cases}\sqrt{4x+1}=a\\\sqrt{3x-2}=b\end{cases}\Rightarrow a^2-b^2=4x+1-3x+2=x+3}\)
Phương trình \(\Leftrightarrow a-b=\frac{a^2-b^2}{5}\Leftrightarrow\left(a-b\right)\left(1-\frac{a+b}{5}\right)=0\Leftrightarrow\orbr{\begin{cases}a=b\\1-\frac{a+b}{5}=0\end{cases}}\)
Với \(a=b\Rightarrow\sqrt{4x+1}=\sqrt{3x-2}\Leftrightarrow4x+1=3x-2\Leftrightarrow x=-3\left(l\right)\)
Với \(\frac{a+b}{5}=1\Rightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\Leftrightarrow7x-1+2\sqrt{12x^2-5x-2}=25\)
\(\Leftrightarrow4\left(12x^2-5x-2\right)=676-364x+49x^2\Leftrightarrow-x^2+344x-684=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=342\left(l\right)\end{cases}}\)vì \(\sqrt{4.342+1}-\sqrt{3.342-2}=5\ne\frac{342+3}{5}=69\)
Vậy pt có nghiệm \(x=2\)
