Giá trị nào của $x$ thỏa mãn \(\dfrac{3}{{1 - 2x}} = \dfrac{{ - 5}}{{3x - 2}}\)

${x^{18}}:{x^6}(x\; \ne 0)$

${x^4}.{\rm{ }}{x^8}$

${x^2}.{\rm{ }}{x^6}$

${\left( {{x^3}} \right)^4}$

$\left| {-{\rm{ }}0,4} \right|{\rm{ }} = {\rm{ }}0,4$

$\left| {-{\rm{ }}0,4} \right|{\rm{ }} = -{\rm{ }}0,4$

$\left| {-\,0,4} \right| =  \pm {\rm{ }}0,4$

$\left| {-{\rm{ }}0,4} \right| = 0$

$\dfrac{{22}}{{15}}$

$\dfrac{6}{8}$

$\dfrac{6}{{15}}$

$\dfrac{8}{{15}}$

\(17\)

\(27\)

\(135\)

\(35\)

\(x = 0\)

$x =  \pm \;\dfrac{1}{2}$

$x = \;\dfrac{1}{2}$

\(x =  - \dfrac{1}{2}\)

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 27}}{{15}}$

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{9}$

$\dfrac{{15}}{5} = \dfrac{9}{{27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

\(27;\,33\)

\(33;27\)

\(27;44\)

\(27;34\)