Question

a)  \(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)

Answer 1

\(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)

\(\Rightarrow\dfrac{2.\left(3x^2+6x-x-2\right)-3\left(2x^2+1\right)-11.3}{6}=0\)

\(\Rightarrow6x^2+12x-2x-4-6x^2-3-33=0\)

\(\Rightarrow10x=40\Rightarrow x=4\)

Answer 2

PT<=>\(\dfrac{2\left(3x-1\right)\left(x+2\right)}{6}-\dfrac{3\left(2x^2+1\right)}{6}=\dfrac{11}{2}\)

    <=>\(\dfrac{10x-7}{6}=\dfrac{11}{2}\)

    <=>2(10x-7)=66

   <=>10x-7=33

    <=>x=4

 Vậy tập nghiệm của pt là S={4}