\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
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\(=-\frac{3}{4}+\frac{6}{5}-\left(\frac{-1}{8}\right)\))
=\(-\frac{3}{4}+\frac{6}{5}+\frac{1}{8}\)
=\(\frac{9}{20}+\frac{1}{8}\)
=\(\frac{23}{40}\)
Vậy..........
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{6}{5}+\frac{1}{8}\)
= \(\frac{30}{40}-\frac{60}{40}+\frac{48}{40}+\frac{5}{40}\)
= \(\frac{23}{40}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
\(=\frac{75}{100}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\left(-\frac{1}{3}\right)^2\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{-3}{4}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{9}{20}-\frac{1}{9}\)
\(=\frac{61}{180}\)
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\frac{-1}{8}\)
= \(\frac{3}{4}-\frac{6}{4}+\frac{1}{2}\cdot\frac{12}{5}-\frac{-1}{8}\)
= \(-\frac{3}{4}+\frac{6}{5}-\frac{-1}{8}=\frac{-30}{40}+\frac{48}{40}+\frac{-6}{40}=\frac{12}{40}=\frac{3}{10}\)
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