\(pt\Leftrightarrow\sqrt{5x-6}-2+\sqrt{10-3x}-2-2x^2+x+6=0\)
\(\Leftrightarrow\frac{5x-10}{\sqrt{5x-6}+2}+\frac{6-3x}{\sqrt{10-3x}+2}-\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-6}+2}+\frac{3}{\sqrt{10-3x}+2}-2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\frac{5}{\sqrt{5x-6}+2}+\frac{3}{\sqrt{10-3x}+2}-2x-3=0\end{matrix}\right.\)
\(\frac{5}{\sqrt{5x-6}+2}+\frac{3}{\sqrt{10-3x}+2}-2x-3=0\)
Ta thấy \(\left\{{}\begin{matrix}x\ge\frac{6}{5}\\x\le\frac{10}{3}\end{matrix}\right.\)(vô lý)
Vậy pt có nghiệm duy nhất x=2
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