Tìm \(x,\) biết:

a) \(x - \frac{3}{8} = \frac{1}{4}.\)                

b) \(x + \frac{2}{3} = \frac{4}{{27}}.\)

c) \(\frac{7}{{15}} + \left( {\frac{5}{6} - x} \right) = \frac{9}{{10}}.\) 

d) \(1,3x - 2,5 = 3,5.\)

e) \(0,2 + 0,8:x = 0,15.\)                                    

f) \(\frac{1}{3}:\left( {2x - 1} \right) = \frac{{ - 4}}{{21}}.\)

g) \(\frac{2}{5} - \frac{x}{7} = 25\%  + \frac{2}{{ - 9}}.\)          

h) \(60\% x + \frac{1}{5}x = \frac{4}{{25}}.\)

i) \(\frac{1}{2}\left( {x - \frac{2}{3}} \right) - \frac{1}{3}\left( {2x - 3} \right) = x.\)   

j) \({\left( {x + \frac{3}{5}} \right)^2} - \frac{9}{{25}} = 0.\)

k) \(\left( {3x - 1} \right)\left( { - \frac{1}{2}x + 5} \right) = 0.\) 

l) \({\left( {3x - \frac{1}{2}} \right)^3} - \frac{1}{{27}} = 0.\)

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 \(x \in \left\{ {\frac{1}{3};\,\,10} \right\}.\)