Giải các phương trình sau:
a) \(7 - \left( {2x + 4} \right) = - \left( {x + 4} \right)\)
b) \(\frac{{1 - 3x}}{6} + x - 1 = \frac{{x + 2}}{2}\)
c) \(\frac{{8x - 3}}{4} - \frac{{3x - 2}}{2} = \frac{{2x - 1}}{2} + \frac{{x + 3}}{4}\)
Đưa phương trình về dạng \(ax + b = 0\) để giải.
a) \(7 - \left( {2x + 4} \right) = - \left( {x + 4} \right)\)
\(\begin{array}{l}7 - 2x - 4 = - x - 4\\ - 2x + x = - 4 - 7 + 4\\ - x = - 7\\x = 7\end{array}\)
Vậy \(x = 7\)
b) \(\frac{{1 - 3x}}{6} + x - 1 = \frac{{x + 2}}{2}\)
\(\begin{array}{l}\frac{{1 - 3x}}{6} + \frac{{6\left( {x - 1} \right)}}{6} = \frac{{3\left( {x + 2} \right)}}{6}\\1 - 3x + 6x - 6 = 3x + 6\\ - 3x + 6x - 3x = 6 + 6 - 1\end{array}\)
\(0 = 11\) (vô lý)
Vậy phương trình vô nghiệm.
c) \(\frac{{8x - 3}}{4} - \frac{{3x - 2}}{2} = \frac{{2x - 1}}{2} + \frac{{x + 3}}{4}\)
\(\begin{array}{l}\frac{{8x - 3}}{4} - \frac{{x + 3}}{4} = \frac{{2x - 1}}{2} + \frac{{3x - 2}}{2}\\\frac{{8x - 3 - x - 3}}{4} = \frac{{2x - 1 + 3x - 2}}{2}\\\frac{{7x - 6}}{4} = \frac{{5x - 3}}{2}\\\frac{{7x - 6}}{4} = \frac{{2\left( {5x - 3} \right)}}{4}\\7x - 6 = 10x - 6\\7x - 10x = - 6 + 6\\ - 3x = 0\\x = 0\end{array}\)
Vậy \(x = 0\).
