ĐKXĐ: \(\dfrac{2}{3}\le x\le2\)
Ta có:
\(VT=\left(x+1\right).2.1\sqrt{3x-2}+\left(2x-1\right).2.1.\sqrt{2-x}\)
\(VT\le\left(x+1\right)\left(1+3x-2\right)+\left(2x-1\right)\left(1+2-x\right)\)
\(VT\le x^2+9x-4\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}3x-2=1\\2-x=1\end{matrix}\right.\) \(\Leftrightarrow x=1\)


