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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}-\frac{3}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{6}{5}\)
\(=-8+\left(\frac{3}{2}-\frac{3}{2}\right)+1+\frac{6}{5}\)
\(=-7+\frac{6}{5}\)
\(=\frac{-35}{5}+\frac{6}{5}=-\frac{29}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Lm r nkoa!
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(=\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)
\(\Rightarrow x=20:0,25=80\)
\(\Rightarrow x=80\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(=\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2}{5}:0,1=4\)
\(\Rightarrow4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\frac{\left(1,2\cdot0,5\right)}{\frac{4}{5}}\)
\(=\frac{\frac{4}{5}:\left(\frac{4}{5}\cdot\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right)\cdot\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right)\cdot\frac{36}{17}}+\frac{6}{5}\cdot\frac{1}{2}\cdot\frac{5}{4}\)
\(=\frac{\frac{4}{5}}{\frac{3}{5}}+\frac{\frac{7}{4}}{\frac{119}{36}\cdot\frac{36}{17}}+\frac{3}{4}\)
\(=\frac{4}{5}\cdot\frac{5}{3}+\frac{7}{4}\cdot\frac{1}{7}+\frac{3}{4}=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}=\frac{7}{3}\)
\(\left(1+\frac{2}{3}-\frac{1}{4}\right)\left(0.8-\frac{3}{4}\right)^2\)
= \(\left(\frac{12+8-3}{12}\right)\left(\frac{3,2-3}{4}\right)^2\)
= \(\frac{17}{12}.\left(\frac{0,2}{4}\right)^2\)
\(\frac{17}{12}.\frac{1}{400}\)
=\(\frac{17}{4800}\)
\(\left(1+\frac{2}{3}-\frac{1}{4}\right)\left(0,8-\frac{3}{4}\right)^2\)
\(=\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right)\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}\cdot\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
\(=\frac{17}{12}\cdot\left(\frac{1}{20}\right)^2\)
\(=\frac{17}{12}\cdot\frac{1}{400}\)
\(=\frac{17}{4800}\)