Tìm x , biết: a) − 1 2/5 x = 3/5 − 8/3 ;

a) \( - 1\frac{2}{5}x = \frac{3}{5} - \frac{8}{3}\)

\(\frac{{ - 7}}{5}x = \frac{{ - 31}}{{15}}\)

\(x = \frac{{ - 31}}{{15}}:\frac{{ - 7}}{5}\)

\(x = \frac{{31}}{{15}}\,\,.\,\,\frac{5}{7}\)

\(x = \frac{{31}}{{21}}\)

Vậy \(x = \frac{{31}}{{21}}\).

b) \({\left( {\sqrt {\frac{{64}}{{25}}} + x} \right)^2} - \frac{{23}}{{18}} = - \frac{7}{{12}}\)

\({\left( {\frac{8}{5} + x} \right)^2} - \frac{{23}}{{18}} = - \frac{7}{{12}}\)

\({\left( {\frac{8}{5} + x} \right)^2} = - \frac{7}{{12}} + \frac{{23}}{{18}}\)

\({\left( {\frac{8}{5} + x} \right)^2} = \frac{{25}}{{36}}\)

\({\left( {\frac{8}{5} + x} \right)^2} = {\left( {\frac{5}{6}} \right)^2}\)

TH1: \(\frac{8}{5} + x = \frac{5}{6}\)

\(x = \frac{5}{6} - \frac{8}{5}\)

\(x = \frac{{ - 23}}{{30}}\)

TH2: \(\frac{8}{5} + x = \frac{{ - 5}}{6}\)

\(x = \frac{{ - 5}}{6} - \frac{8}{5}\)

\(x = \frac{{ - 73}}{{30}}\).

Vậy \(x \in \left\{ {\frac{{ - 23}}{{30}};\,\,\frac{{ - 73}}{{30}}} \right\}\).

c) \[2\,\,.\,\,{5^{x + 1}} + 4\,\,.\,\,{5^{x + 2}} = 22\,\,.\,\,{5^3}\]

\[2\,\,.\,\,{5^{x + 1}} + 4\,\,.\,5\,.\,\,{5^{x + 1}} = 22\,\,.\,\,{5^3}\]

\[2\,\,.\,\,{5^{x + 1}} + 20\,.\,\,{5^{x + 1}} = 22\,\,.\,\,{5^3}\]

\[(2 + 20)\,.\,\,{5^{x + 1}} = 22\,\,.\,\,{5^3}\]

\[22\,.\,\,{5^{x + 1}} = 22\,\,.\,\,{5^3}\]

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

\[x + 1 = 3\]

\[x = 2\]

Vậy \[x = 2\].