a) \(2\,x\, - \,\frac{5}{4}\, = \,\frac{{20}}{{15}}\)

\(2x = \frac{4}{3} + \frac{5}{4} = \frac{{31}}{{12}}\)

\(x\, = \frac{{31}}{{12}}\,:2 = \frac{{31}}{{24}}\).

Vậy \(x\, = \frac{{31}}{{24}}\).

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

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

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

\(x = \frac{{ - \,1}}{2} - \frac{1}{5}\,\, = \,\,\frac{{ - \,7}}{{10}}\).

Vậy \(x\, = \,\,\frac{{ - \,7}}{{10}}\)

c) \(\frac{{ - 2}}{3}x + 1,6\, = \,\frac{3}{2}\)

\(\frac{{ - 2}}{3}x\, = \,\frac{3}{2} + \frac{{16}}{{10}}\)

\(\frac{{ - 2}}{3}x\, = \,\frac{3}{2} + \frac{8}{5} = \frac{{31}}{{10}}\)

\(x = \frac{{ - 93}}{{20}}\)

Vậy \(x = \frac{{ - 93}}{{20}}\)

d) \(\frac{{243}}{{{3^{x + 1}}}} = {3^x}\)

\[243 = {3^x}\,.\,{3^x}{^ + ^1}\]

\[{3^5} = {3^x}^{ + x + 1}\]

\[{3^5} = {3^2}^{x + 1}\]

\[2x + 1 = 5\]

\[x = 2\].

Vậy \[x = 2\].

Làm tròn số 12,7362 đến chữ số thập phân thứ hai là