\(\left(1+\frac{2}{3}-\frac{1}{4}\right)\left(0,8-\frac{3}{4}\right)^2\)
Giải giúp mình nhé, thank!
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\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}-\frac{3}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{6}{5}\)
\(=-8+\left(\frac{3}{2}-\frac{3}{2}\right)+1+\frac{6}{5}\)
\(=-7+\frac{6}{5}\)
\(=\frac{-35}{5}+\frac{6}{5}=-\frac{29}{5}\)
=\(\frac{-16}{5}.\frac{-15}{64}+\left(\frac{4}{5}-\frac{34}{15}\right):\frac{7}{2}\)
= \(\frac{-16.\left(-15\right)}{5.64}+\left(\frac{12}{15}-\frac{34}{15}\right).\frac{2}{7}\)
=\(\frac{-1.\left(-3\right)}{1.4}+\left(\frac{-22}{15}\right).\frac{2}{7}\)
=\(\frac{3}{4}-\frac{44}{105}=\frac{315}{420}-\frac{176}{420}=\frac{139}{420}\)
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)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
\(\left(1+\frac{2}{3}-\frac{1}{4}\right)\left(0.8-\frac{3}{4}\right)^2\)
= \(\left(\frac{12+8-3}{12}\right)\left(\frac{3,2-3}{4}\right)^2\)
= \(\frac{17}{12}.\left(\frac{0,2}{4}\right)^2\)
\(\frac{17}{12}.\frac{1}{400}\)
=\(\frac{17}{4800}\)
\(\left(1+\frac{2}{3}-\frac{1}{4}\right)\left(0,8-\frac{3}{4}\right)^2\)
\(=\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right)\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}\cdot\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
\(=\frac{17}{12}\cdot\left(\frac{1}{20}\right)^2\)
\(=\frac{17}{12}\cdot\frac{1}{400}\)
\(=\frac{17}{4800}\)