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a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
\(2^3+3.\left(\frac{2}{3}\right)^0-2+\left[\left(-2\right)^2:\frac{1}{2}\right]-8\)
đổi p/s \(\left(\frac{2}{3}\right)^0=1\)
xong tính trong ngoặc vuông,
r xử dụng tính chất phân phối
= -19
_______
8
chúc bạn học tốt !