Tìm x biết: câu a) 2/3 + x = -1/12; câu b) (2x-4/3) (1+6/5 x) = 0
a) \(\frac{2}{3} + x = \frac{{ - 1}}{{12}}\)
\(x = \frac{{ - 1}}{{12}} - \frac{2}{3}\)
\(x = \frac{{ - 9}}{{12}}\)
\(x = \frac{{ - 3}}{4}\)
Vậy \(x = \frac{{ - 3}}{4}\).
b) \(\left( {2x - \frac{4}{3}} \right)\left( {1 + \frac{6}{5}x} \right) = 0\)
Suy ra \(2x - \frac{4}{3} = 0\) hoặc \(1 + \frac{6}{5}x = 0\)
\(2x = \frac{4}{3}\) \(\frac{6}{5}x = - 1\)
\(x = \frac{4}{3}:2\) \(x = - 1:\frac{6}{5}\)
\(x = \frac{2}{3}\) \(x = - \frac{5}{6}\)
Vậy \[x \in \left\{ {\frac{2}{3}; - \frac{5}{6}} \right\}\].
c) \[2\left| {x - \frac{1}{3}} \right| + \frac{7}{4} = {\left( { - \frac{3}{2}} \right)^6}:{\left( { - \frac{3}{2}} \right)^4}\]
\[2\left| {x - \frac{1}{3}} \right| = {\left( { - \frac{3}{2}} \right)^2} - \frac{7}{4}\]
\[2\left| {x - \frac{1}{3}} \right| = \frac{9}{4} - \frac{7}{4}\]
\[2\left| {x - \frac{1}{3}} \right| = \frac{1}{2}\]
\[\left| {x - \frac{1}{3}} \right| = \frac{1}{4}\]
Trường hợp 1: \[x - \frac{1}{3} = \frac{1}{4}\] \[x = \frac{1}{4} + \frac{1}{3}\] \[x = \frac{7}{{12}}\] | Trường hợp 2: \[x - \frac{1}{3} = - \frac{1}{4}\] \[x = - \frac{1}{4} + \frac{1}{3}\] \[x = \frac{1}{{12}}\] |
Vậy \(x \in \left\{ {\frac{7}{{12}};\frac{1}{{12}}} \right\}\).