Giải các phương trình sau:
a) \(8 + 2\left( {x - 1} \right) = 20\)
b) \(4\left( {3x - 2} \right) + 3\left( {x - 4} \right) = 7x + 20\)
c) \(\frac{{2x}}{3} + x = \frac{{2x + 5}}{6} + \frac{1}{2}\)
Đưa phương trình về dạng \(ax + b = 0\) để giải.
a) \(8 + 2\left( {x - 1} \right) = 20\)
\(\begin{array}{l}8 + 2x - 2 = 20\\2x + 6 = 20\\2x = 20 - 6\\2x = 14\\x = 7\end{array}\)
Vậy \(x = 7\)
b) \(4\left( {3x - 2} \right) + 3\left( {x - 4} \right) = 7x + 20\)
\(\begin{array}{l}12x - 8 + 3x - 12 = 7x + 20\\12x + 3x - 7x = 20 + 8 + 12\\8x = 40\\x = 5\end{array}\)
Vậy \(x = 5\)
c) \(\frac{{2x}}{3} + x = \frac{{2x + 5}}{6} + \frac{1}{2}\)
\(\begin{array}{l}\frac{{2.2x}}{6} + \frac{{6x}}{6} = \frac{{2x + 5}}{6} + \frac{3}{6}\\4x + 6x = 2x + 5 + 3\\10x - 2x = 8\\8x = 8\\x = 1\end{array}\)
Vậy \(x = 1\)