Điều kiện: \(\left\{{}\begin{matrix}3x-2\ge0\\7-x\ge0\end{matrix}\right.\Leftrightarrow\dfrac{2}{3}\le x\le7\)
\(\sqrt{3x-2}-\sqrt{7-x}+3x^2-20x=-9\)
\(\Leftrightarrow\left(\sqrt{3x-2}-4\right)-\left(\sqrt{7-x}-1\right)+\left(3x^2-20x+12\right)=0\)
\(\Leftrightarrow\dfrac{\left(3x-2\right)-16}{\sqrt{3x-2}+4}-\dfrac{\left(7-x\right)-1}{\sqrt{7-x}+1}+\left[\left(3x^2-18x\right)-\left(2x-12\right)\right]=0\)
\(\Leftrightarrow\dfrac{3x-18}{\sqrt{3x-2}+4}+\dfrac{x-6}{\sqrt{7-x}+1}+\left[3x\left(x-6\right)-2\left(x-6\right)\right]=0\)
\(\Leftrightarrow\dfrac{3\left(x-6\right)}{\sqrt{3x-2}+4}+\dfrac{x-6}{\sqrt{7-x}+1}+\left(3x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left[\dfrac{3}{\sqrt{3x-2}+4}+\dfrac{1}{\sqrt{7-x}+1}+\left(3x-2\right)\right]=0\)
\(\Leftrightarrow x-6=0\) (do \(\dfrac{3}{\sqrt{3x-2}+4}+\dfrac{1}{\sqrt{7-x}+1}+\left(3x-2\right)>0\))
\(\Leftrightarrow x=6\) (thỏa mãn)
Vậy phương trình có tập nghiệm là: \(S=\left\{6\right\}.\)
