Tìm x ( x thuộc Z) biết: x - 3/8 = 1/4
Hướng dẫn giải:
a) \(x - \frac{3}{8} = \frac{1}{4}\) \(x = \frac{1}{4} + \frac{3}{8}\) \(x = \frac{5}{8}\) Vậy \(x = \frac{5}{8}.\) c) \(\frac{7}{{15}} + \left( {\frac{5}{6} - x} \right) = \frac{9}{{10}}\) \(\left( {\frac{5}{6} - x} \right) = \frac{9}{{10}} - \frac{7}{{15}}\) \(\frac{5}{6} - x = \frac{{13}}{{30}}\) \(x = \frac{5}{6} - \frac{{13}}{{30}}\) \(x = \frac{2}{5}.\) Vậy \(x = \frac{2}{5}.\) e) \(0,2 + 0,8:x = 0,15\) \(0,8:x = 0,15 - 0,2\) \(0,8:x = - 0,05\) \(x = 0,8:\left( { - 0,05} \right)\) \(x = - 16\) Vậy \(x = - 16.\) g) \(\frac{2}{5} - \frac{x}{7} = 25\% + \frac{2}{{ - 9}}\) \(\frac{2}{5} - \frac{x}{7} = \frac{1}{4} - \frac{2}{9}\) \(\frac{2}{5} - \frac{x}{7} = \frac{1}{{36}}\) \(\frac{x}{7} = \frac{2}{5} - \frac{1}{{36}}\) \(\frac{x}{7} = \frac{{67}}{{180}}\) \(x = \frac{{469}}{{180}}.\) Vậy \(x = \frac{{469}}{{180}}.\) i) \(\frac{1}{2}\left( {x - \frac{2}{3}} \right) - \frac{1}{3}\left( {2x - 3} \right) = x\) \(\frac{1}{2}x - \frac{1}{3} - \frac{2}{3}x + 1 - x = 0\) \(\left( {\frac{1}{2}x - \frac{2}{3}x - x} \right) + \left( { - \frac{1}{3} + 1} \right) = 0\) \(\left( {\frac{1}{2} - \frac{2}{3} - 1} \right)x + \frac{2}{3} = 0\) \(\frac{{ - 7}}{6}x + \frac{2}{3} = 0\) \(\frac{{ - 7}}{6}x = - \frac{2}{3}\) \(x = - \frac{2}{3}:\frac{{ - 7}}{6}\) \(x = - \frac{2}{3} \cdot \frac{{ - 6}}{7}\) \(x = \frac{4}{7}\) Vậy \(x = \frac{4}{7}.\) k) \(\left( {3x - 1} \right)\left( { - \frac{1}{2}x + 5} \right) = 0.\) | b) \(x + \frac{2}{3} = \frac{4}{{27}}\) \(x = \frac{4}{{27}} - \frac{2}{3}\) \(x = \frac{{ - 14}}{{27}}\) Vậy \(x = \frac{{ - 14}}{{27}}.\) d) \(1,3x - 2,5 = 3,5\) \(1,3x = 3,5 + 2,5\) \(1,2x = 6\) \(x = 6:1,3\) \(x = \frac{{60}}{{13}}.\) Vậy \(x = \frac{{60}}{{13}}.\) f) \(\frac{1}{3}:\left( {2x - 1} \right) = \frac{{ - 4}}{{21}}\) \(2x - 1 = \frac{1}{3}:\frac{{ - 4}}{{21}}\) \(2x - 1 = \frac{1}{3} \cdot \frac{{ - 21}}{4}\) \(2x - 1 = \frac{{ - 7}}{4}\) \(2x = \frac{{ - 7}}{4} + 1\) \(2x = \frac{{ - 7}}{4} + \frac{4}{4}\) \(2x = \frac{{ - 3}}{4}\) \(x = \frac{{ - 3}}{4}:2\) \(x = \frac{{ - 3}}{{4 \cdot 2}}\) \(x = \frac{{ - 3}}{8}.\) Vậy \(x = \frac{{ - 3}}{8}.\) h) \(60\% x + \frac{1}{5}x = \frac{4}{{25}}\) \(\frac{3}{5}x + \frac{1}{5}x = \frac{4}{{25}}\) \(\left( {\frac{3}{5} + \frac{1}{5}} \right)x = \frac{4}{{25}}\) \(\frac{4}{5}x = \frac{4}{{25}}\) \(x = \frac{4}{{25}}:\frac{4}{5}\) \(x = \frac{1}{5}.\) Vậy \(x = \frac{1}{5}.\) j) \({\left( {x + \frac{3}{5}} \right)^2} - \frac{9}{{25}} = 0.\) \({\left( {x + \frac{3}{5}} \right)^2} - \frac{9}{{25}} = 0\) \({\left( {x + \frac{3}{5}} \right)^2} = \frac{9}{{25}}\) | ||
Trường hợp 1: \(x + \frac{3}{5} = \frac{3}{5}\) \(x = \frac{3}{5} - \frac{3}{5}\) \(x = 0;\) | Trường hợp 2: \(x + \frac{3}{5} = - \frac{3}{5}\) \(x = - \frac{3}{5} - \frac{3}{5}\) \(x = \frac{{ - 6}}{5}.\) | ||
Trường hợp 1: | Trường hợp 2: | ||
\(3x - 1 = 0\) \(3x = 1\) \(x = \frac{1}{3};\) | \( - \frac{1}{2}x + 5 = 0\)\( - \frac{1}{2}x = - 5\) \(x = 10.\) | Vậy \(x \in \left\{ {0;\,\,\frac{{ - 6}}{5}} \right\}.\) l) \({\left( {3x - \frac{1}{2}} \right)^3} - \frac{1}{{27}} = 0.\) \({\left( {3x - \frac{1}{2}} \right)^3} = {\left( {\frac{1}{3}} \right)^3}\) \(3x - \frac{1}{2} = \frac{1}{3}\) \(3x = \frac{5}{6}\) \(x = \frac{5}{{18}}\) Vậy \(x = \frac{5}{{18}}.\) | |
Vậy \(x \in \left\{ {\frac{1}{3};\,\,10} \right\}.\) | |||