tính tổng
\(\frac{3}{7}+\frac{-2}{3}\)
tù đó có thể suy ra ngay kết quả các phép ccongj sau ko
\(\frac{9}{21}+\frac{-10}{15};\frac{24}{56}+\frac{-18}{27}\)
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a) Ta có: BCNN(15,10) = 30 nên ta chọn mẫu số chung là 30
\(\frac{11}{15}+\frac{9}{10}=\frac{22}{30}+\frac{27}{30}=\frac{49}{30}\)
b) Ta có: BCNN(6,9,12) = 36 nên ta chọn mẫu số chung là 36
\(\frac{5}{6} + \frac{7}{9} + \frac{{11}}{{12}} = \frac{{30}}{{36}} + \frac{{28}}{{36}} + \frac{{33}}{{36}} = \frac{{91}}{{36}}\)
c) Ta có: BCNN(24,21) = 168 nên ta chọn mẫu số chung là 168
\(\frac{7}{{24}} - \frac{2}{{21}} = \frac{{49}}{{168}} - \frac{{16}}{{168}} = \frac{{33}}{{168}}=\frac{11}{56}\)
d) Ta có: BCNN(36,24) = 72 nên ta chọn mẫu số chung là 72
\(\frac{{11}}{{36}} - \frac{7}{{24}} = \frac{{22}}{{72}} - \frac{{21}}{{72}} = \frac{1}{{72}}\)
a)
\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).
Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)
b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)
c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)
d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
Ta có: \(\frac{9}{21}=\frac{9:3}{21:3}=\frac{3}{7}\)
Và \(\frac{-10}{15}=\frac{-10:5}{15:5}=-\frac{2}{3}\)
\(\Rightarrow\frac{9}{21}+\frac{-10}{15}=\frac{3}{7}+\frac{-2}{3}=\frac{9}{21}+\frac{-14}{21}=\frac{9+\left(-14\right)}{21}=\frac{-5}{21}\)
Tương tự: \(\frac{24}{56}=\frac{24:8}{56:8}=\frac{3}{7}\)
Và \(\frac{-18}{27}=\frac{-18:9}{27:9}=\frac{-2}{3}\)
\(\Rightarrow\frac{24}{56}+\frac{-18}{27}=\frac{3}{7}+\frac{-2}{3}=\frac{-5}{21}\)
\(\frac{3}{7}+\frac{-2}{3}=\frac{9}{21}-\frac{14}{21}=\frac{-5}{21}\)
\(\frac{9}{21}+\frac{-10}{15}=\frac{3}{7}-\frac{2}{3}=\frac{-5}{21}\)
\(\frac{24}{56}+\frac{-18}{27}=\frac{3}{7}-\frac{2}{3}=\frac{-5}{21}\)