ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(PT\Leftrightarrow\sqrt{3x-2}-\sqrt{x+1}=\left(x+1\right)\left(2x-3\right)\)
Ta thấy \(2x-3=3x-2-\left(x+1\right)\)
Đặt\(\left\{{}\begin{matrix}\sqrt{3x-2}=a\\\sqrt{x+1}=b\end{matrix}\right.\left(a,b\ne0\right)\)
\(PT\Leftrightarrow a-b=b^2.\left(a^2-b^2\right)\)
\(\Leftrightarrow\left(a-b\right)-b^2\left(a-b\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(ab^2+b^3-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\ab^2+b^3-1=0\end{matrix}\right.\)
Vì \(ab^2+b^3-1=0\) vô nghiệm
\(\Rightarrow a=b\)
\(\Rightarrow\sqrt{3x-2}=\sqrt{x+1}\)
\(\Leftrightarrow3x-2=x+1\Rightarrow x=\dfrac{3}{2}\)(t/m)
Vậy...
