Hình nào dưới đây biểu diễn hai góc kè bù?

Ta có: \({4^{30}} = {2^{30}}{.2^{30}} = {\left( {{2^3}} \right)^{10}}.{\left( {{2^2}} \right)^{15}} = {8^{10}}{.4^{15}}\).

Mà \({8^{10}}\,\,.\,\,{4^{15}} > {8^{10}}\,\,.\,\,{3^{15}} > {8^{10}}\,\,.\,\,{3^{11}} > {8^{10}}\,\,.\,\,{3^{10}}\,\,.\,\,3 = {\left( {8\,\,.\,\,3} \right)^{10}}\,\,.\,\,3 = {24^{10}}\,\,.\,\,3\).

Vì \[{4^{30}} > 3\,\,.\,\,{24^{10}}\] nên \({2^{30}} + {3^{30}} + {4^{30}} > 3\,\,.\,\,{24^{10}}\).

1. a) \(\frac{{ - 5}}{{17}}.\frac{{31}}{{33}} + \frac{{ - 5}}{{17}}.\frac{2}{{33}} + 2\frac{5}{{17}}\)\( = \frac{{ - 5}}{{17}}.\left( {\frac{{31}}{{33}} + \frac{2}{{33}}} \right) + 2\frac{5}{{17}}\)\( = \frac{{ - 5}}{{17}} + 2\frac{5}{{17}}\)\( = 2\);

b) \(15.{\left( {\frac{2}{3}} \right)^2} + {\left( {\frac{{23}}{6}} \right)^0}.\frac{{24}}{{16}} - 2\frac{2}{3}\)\( = 15.\frac{4}{9} + 1.\frac{{24}}{{16}} - 2\frac{2}{3}\)\( = \frac{{20}}{3} + \frac{3}{2} - \frac{8}{3}\)

\( = \left( {\frac{{20}}{3} - \frac{8}{3}} \right) + \frac{3}{2}\)\( = \frac{{12}}{3} + \frac{3}{2}\)\( = \frac{{11}}{2}\).

2.

a) \(\left( {\frac{1}{2}x - \frac{1}{5}} \right).\frac{{ - 1}}{2} = \frac{3}{4}\)

\(\frac{1}{2}x - \frac{1}{5} = \frac{3}{4}:\frac{{ - 1}}{2}\)

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

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

\(\frac{1}{2}x = \frac{{ - 13}}{{10}}\)

\(x = \frac{{ - 13}}{{10}}:\frac{1}{2}\)

\(x = \frac{{ - 13}}{5}\)

Vậy \(x = \frac{{ - 13}}{5}\).

b) \(\left| {x + \frac{5}{6}} \right| + \frac{7}{2} = 5\)

\(\left| {x + \frac{5}{6}} \right| = 5 - \frac{7}{2}\)

\(\left| {x + \frac{5}{6}} \right| = \frac{3}{2}\)

Trường hợp 1: \(x + \frac{5}{6} = \frac{3}{2}\)

\(x = \frac{3}{2} - \frac{5}{6}\)

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

Trường hợp 2: \(x + \frac{5}{6} = \frac{{ - 3}}{2}\)

\(x = \frac{{ - 3}}{2} - \frac{5}{6}\)

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

Vậy \(x \in \left\{ {\frac{2}{3};\,\,\frac{{ - 7}}{3}} \right\}\).