Question

\((x-3)- \dfrac{(x-3)(2x-5)}{6}=\dfrac{(x-3)(3-x)}{4}\)

Giải pt

Answer 1

\(\Leftrightarrow\dfrac{12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)-3\left(x-3\right)\left(3-x\right)}{12}=0\)

\(\Leftrightarrow12x-36-2\left(2x^2-5x-6x+15\right)-3\left(3x-x^2-9+3x\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30-18x+3x^2+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Delta=b^2-4ac=16^2-4.\left(-1\right).\left(-39\right)=100>0\)

\(\Rightarrow PT\) có 2 nghiệm pb \(x_1,x_2\)

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-16+10}{-2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-16-10}{-2}=13\end{matrix}\right.\)

Vậy \(S=\left\{3;13\right\}\)