=>(2x+5)(3x-2)=55
\(\Leftrightarrow6x^2-4x+15x-10-55=0\)
\(\Leftrightarrow6x^2+11x-65=0\)
\(\text{Δ}=11^2-4\cdot6\cdot\left(-65\right)=1681>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-11-41}{12}=\dfrac{-52}{12}=-\dfrac{13}{3}\\x_2=\dfrac{-11+41}{12}=\dfrac{30}{12}=\dfrac{5}{2}\end{matrix}\right.\)
\(62-\left(2x+5\right)\times\left(3x-2\right)=7\)
\(\Leftrightarrow62-\left(6x^2-4x+15x-10\right)=7\)
\(\Leftrightarrow6x^2+11x-10=62-7\)
\(\Leftrightarrow6x^2+11x-10=55\)
\(\Leftrightarrow6x^2+11x-65=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{13}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{5}{2};-\dfrac{13}{3}\right\}\)
