Question

Giải phương trình sau: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

Answer 1

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow12\left(x-3\right)-2\left(2x^2-11x+15\right)=-3\left(x^2-6x+9\right)\)

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

\(\Leftrightarrow-4x^2+34x-66=-3x^2+18x-27\)

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

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy \(S=\left\{13;3\right\}\)