\(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}\) - \(\dfrac{2x^2+1}{2}\)= \(\dfrac{11}{2}\)
MTC = 6.
\(\Leftrightarrow\) 2(3x-1)(x+2) - 3(2x2+1) = 3.11
\(\Leftrightarrow\) 2(3x2+5x-2) - 6x2-3 = 33
\(\Leftrightarrow\) 6x2 + 10x - 4 - 6x2 - 3 - 33 = 0
\(\Leftrightarrow\) 10x - 40 = 0
\(\Leftrightarrow\) 10x = 40
\(\Leftrightarrow\) x = 4
Vậy S =\(\left\{4\right\}\)