a, 4.(3\(x\) - 2) - 3.(5\(x\) + 6) = 35
12\(x\) - 8 - 15\(x\) - 18 = 35
\(x\)(12 - 15) - ( 8 + 18) = 35
- 3\(x\) - 26 = 35
3\(x\) = - 26 - 35
3\(x\) = - 61
\(x\) = - \(\dfrac{61}{3}\)
c, \(x^2\) - 5\(x\) = 0
\(x\).(\(x-5\)) =0
\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
vậy \(x\) \(\in\) {0; 5}