Tìm \(x\), biết:

d) \(\frac{{{{( - 3)}^x}}}{{81}} = - 27\);

b) \(\left( {1 + \frac{2}{3} - \frac{1}{4}} \right){\left( {0,8 - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{12}}{{12}} + \frac{8}{{12}} - \frac{3}{{12}}} \right){\left( {\frac{4}{5} - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{17}}{{12}}} \right){\left( {\frac{1}{{20}}} \right)^2}\)

\( = \frac{{17}}{{12}}\, \cdot \,\frac{1}{{400}} = \frac{{17}}{{4800}}\).

b) \[\left| {2x + \frac{1}{3}} \right| - \sqrt {1\frac{9}{{16}}} = {\left( { - \frac{{\sqrt 3 }}{2}} \right)^2}\]

\[\left| {2x + \frac{1}{3}} \right| - \sqrt {\frac{{25}}{{16}}} = \frac{3}{4}\]

\[\left| {2x + \frac{1}{3}} \right| - \frac{5}{4} = \frac{3}{4}\]

\[\left| {2x + \frac{1}{3}} \right| = 2\]

TH1: \(2x + \frac{1}{3} = 2\)

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

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

TH2: \(2x + \frac{1}{3} = - 2\)

\(2x = \frac{{ - 7}}{3}\)

\(x = \frac{{ - 7}}{6}\)