Đề bài

Thực hiện phép tính. a)\(\frac{2}{5} + \frac{3}{5}:\left( { - \frac{3}{2}} \right) + \frac{1}{2};\) b)\(2\frac{1}{3} + {\left( { - \frac{1}{3}} \right)^2} - \frac{3}{2};\) c)\(\left( {\frac{7}{8} - 0,25} \right):{\left( {\frac{5}{6} - 0,75} \right)^2};\) d)\(\left( { - 0,75} \right) - \left[ {\left( { - 2} \right) + \frac{3}{2}} \right]:1,5 + \left( {\frac{{ - 5}}{4}} \right)\)

Lời giải chi tiết

a)\(\begin{array}{l}\frac{2}{5} + \frac{3}{5}:\left( { - \frac{3}{2}} \right) + \frac{1}{2}\\ = \frac{2}{5} + \frac{3}{5}.\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}\\ = \frac{2}{5} + \frac{{ - 2}}{5} + \frac{1}{2}\\ =0+ \frac{1}{2}\\= \frac{1}{2}\end{array}\)b)\(\begin{array}{l}2\frac{1}{3} + {\left( { - \frac{1}{3}} \right)^2} - \frac{3}{2}\\ = \frac{7}{3} + \frac{1}{9} - \frac{3}{2}\\ = \frac{{42}}{{18}} + \frac{2}{{18}} - \frac{{27}}{{18}}\\ = \frac{{17}}{{18}}\end{array}\)c)\(\begin{array}{l}\left( {\frac{7}{8} - 0,25} \right):{\left( {\frac{5}{6} - 0,75} \right)^2}\\ = \left( {\frac{7}{8} - \frac{1}{4}} \right):\left( {\frac{5}{6} - \frac{3}{4}} \right)^2\\ = \left( {\frac{7}{8} - \frac{2}{8}} \right):\left( {\frac{{10}}{{12}} - \frac{9}{{12}}} \right)^2\\ = \frac{5}{8}:(\frac{1}{{12}})^2\\ =\frac{5}{8}:\frac{1}{144}\\= \frac{5}{8}.144\\ = 90\end{array}\)d)\(\begin{array}{l}\left( { - 0,75} \right) - \left[ {\left( { - 2} \right) + \frac{3}{2}} \right]:1,5 + \left( {\frac{{ - 5}}{4}} \right)\\ = \left( {\frac{{ - 75}}{100}} \right) - \left[ {-2 + \frac{3}{2}} \right]:\frac{15}{10} + \left( {\frac{{ - 5}}{4}} \right)\\= \left( {\frac{{ - 3}}{4}} \right) - \left[ {\frac{{ - 4}}{2} + \frac{3}{2}} \right]:\frac{3}{2} + \left( {\frac{{ - 5}}{4}} \right)\\ = \left( {\frac{{ - 3}}{4}} \right) - (\frac{{ - 1}}{2}).\frac{2}{3} + \left( {\frac{{ - 5}}{4}} \right)\\ = \left( {\frac{{ - 3}}{4}} \right) + {\frac{{ 1}}{3}} + \frac{-5}{4}\\= \left( {\frac{{ - 3}}{4}} \right) + \left( {\frac{{ - 5}}{4}} \right) + \frac{1}{3}\\ = \frac{-8}{4} + \frac{1}{3}\\= - 2 + \frac{1}{3}\\ = \frac{{ - 6}}{3} + \frac{1}{3}\\ = \frac{{ - 5}}{3}\end{array}\)