Tìm \(x\), biết:

    a) \( - 0,6 - x = 0,1\);                                   b) \(\frac{2}{3}:x = \frac{8}{5}\);

    c) \(27{\left( {3x - \frac{1}{5}} \right)^3} =  - 8\);        d) \(\left( {x - \frac{4}{5}} \right)\left( {x + 2\frac{1}{5}} \right) = 0\).

a) \( - 0,6 - x = 0,1\)

\(x =  - 0,6 - 0,1\)

\(x =  - 0,7\)

Vậy \(x =  - 0,7\).

c) \(27{\left( {3x - \frac{1}{5}} \right)^3} =  - 8\)

\({\left( {3x - \frac{1}{5}} \right)^3} =  - \frac{8}{{27}} = {\left( { - \frac{2}{3}} \right)^3}\)

Suy ra \(3x - \frac{1}{5} =  - \frac{2}{3}\)

\(3x =  - \frac{2}{3} + \frac{1}{5}\)

\(3x =  - \frac{7}{{15}}\)

\(x =  - \frac{7}{{45}}\)

Vậy \(x =  - \frac{7}{{45}}\).

b) \(\frac{2}{3}:x = \frac{8}{5}\)

\(x = \frac{2}{3}:\frac{8}{5}\)

\(x = \frac{2}{3}.\frac{5}{8}\)

\(x = \frac{5}{{12}}\)

Vậy \(x = \frac{5}{{12}}\).

d) \(\left( {x - \frac{4}{5}} \right)\left( {x + 2\frac{1}{5}} \right) = 0\)

Suy ra \(x - \frac{4}{5} = 0\) hoặc \(x + 2\frac{1}{5} = 0\)

Trường hợp 1:

\(x - \frac{4}{5} = 0\)

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

Vậy \(x \in \left\{ {\frac{4}{5}; - \frac{{11}}{5}} \right\}\).

Trường hợp 2:

\(x + 2\frac{1}{5} = 0\)

\(x =  - 2\frac{1}{5}\)

\(x =  - \frac{{11}}{5}\)