Tìm x , biết: a) 1 /3 + 2/ 3 : x = − 7 ; b) x + 1/ 2 = (x − 1)/ 3 ;
a) \(\frac{1}{3} + \frac{2}{3}:x = - 7\) \(\frac{2}{3}:x = - 7 - \frac{1}{3}\) \(\frac{2}{3}:x = - \frac{{22}}{3}\) \(x = \frac{2}{3}:\left( { - \frac{{22}}{3}} \right)\) \(x = \frac{2}{3}.\left( { - \frac{3}{{22}}} \right)\) b) \(\frac{{x + 1}}{2} = \frac{{x - 1}}{3}\) \(3\left( {x + 1} \right) = 2\left( {x - 1} \right)\) \(3x + 3 = 2x - 2\) \(3x - 2x = - 2 - 3\) \(x = - 5\) Vậy \(x = - 5\). | c) \(\frac{1}{4} + \left| {3x - 1\frac{1}{4}} \right| = \frac{3}{2}\) \(\left| {3x - \frac{5}{4}} \right| = \frac{3}{2} - \frac{1}{4}\) \(\left| {3x - \frac{5}{4}} \right| = \frac{5}{4}\) | |
Trường hợp 1: \(3x - \frac{5}{4} = \frac{5}{4}\) \(3x = \frac{5}{4} + \frac{5}{4}\) \(3x = \frac{5}{2}\) \(x = \frac{5}{2}:3\) \(x = \frac{5}{6}\) Vậy \(x \in \left\{ {\frac{5}{6};0} \right\}\). | Trường hợp 2: \(3x - \frac{5}{4} = - \frac{5}{4}\) \(3x = - \frac{5}{4} + \frac{5}{4}\) \(3x = 0\) \(x = 0\)
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