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

\(a = c\)

\(a.c = b.d\)

\(a.d = b.c\)

\(b = d\)

$\dfrac{5}{{35}} = \dfrac{9}{{63}}$

$\dfrac{{63}}{9} = \dfrac{{35}}{5}$

$\dfrac{{35}}{9} = \dfrac{{63}}{5}$

$\dfrac{{63}}{{35}} = \dfrac{9}{5}$

$\dfrac{7}{{12}}$ và $\dfrac{5}{6}:\dfrac{4}{3}$

$\dfrac{6}{7}:\dfrac{{14}}{5}$ và $\dfrac{7}{3}:\dfrac{2}{9}$

$\dfrac{{15}}{{21}}$ và $-\dfrac{{125}}{{175}}$

$\dfrac{{ - 1}}{3}$  và $\dfrac{{ - 19}}{{57}}$        

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

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

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

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

\(\dfrac{2}{a} = \dfrac{5}{b}\)

\(\dfrac{b}{5} = \dfrac{2}{a}\)

\(\dfrac{2}{5} = \dfrac{a}{b}\)

\(\dfrac{2}{b} = \dfrac{5}{a}\)

\(x = \dfrac{1}{5}\)    

\(x =  - \dfrac{5}{4}\)

\(x = \dfrac{5}{4}\)

\(x = \dfrac{4}{5}\)

\(1\)

\(2\)

\(0\)

\(3\)