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24 tháng 6 2018

\(\frac{8}{7}+\frac{16}{7}+\frac{24}{7}=\frac{48}{7}\)

nếu có sai thông cảm

24 tháng 6 2018

\(1\frac{1}{7}+2\frac{4}{7}+3\frac{3}{7}\)

\(=1+\frac{1}{7}+2+\frac{4}{7}+3+\frac{3}{7}\)

\(=\left(1+2+3\right)+\left(\frac{1}{7}+\frac{4}{7}+\frac{3}{7}\right)\)

\(=6+1\frac{1}{7}\)

\(=6+1+\frac{1}{7}\)

\(=7\frac{1}{7}\)

7 tháng 3 2018

b)  \(\frac{4}{10}+\frac{-2}{9}+\frac{-3}{-5}+\frac{21}{-27}+\frac{-10}{20}\)

\(=\frac{2}{5}-\frac{2}{9}+\frac{3}{5}-\frac{7}{9}-\frac{1}{2}\)

\(=\left(\frac{2}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}+\frac{7}{9}\right)-\frac{1}{2}\)

\(=1-1-\frac{1}{2}=-\frac{1}{2}\)

7 tháng 3 2018

a)  \(-\frac{5}{2}+\frac{1}{7}+\frac{6}{7}+\frac{1}{2}+\frac{3}{4}\)

\(=\left(\frac{-5}{2}+\frac{1}{2}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\frac{3}{4}\)

\(=-2+1+\frac{3}{4}\)

\(=-1+\frac{3}{4}\)

\(=-\frac{1}{4}\)

20 tháng 7 2017

     \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)

\(=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)

\(=\frac{2}{7}-\frac{2}{7}\)

\(=0\)

1 tháng 4 2019

\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)

\(=\frac{2}{7}-\frac{1}{\frac{7}{2}}\)

\(=\frac{2}{7}-\frac{2}{7}=0\)

Study well ! >_<

HQ
Hà Quang Minh
Giáo viên
7 tháng 10 2023

a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)

\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} =  \frac{{ -29}}{{35}}\end{array}\)

b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)

c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)

\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)

d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)

\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ =  (- 2).45,3 =  - 90,6\end{array}\)

a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)

b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)

\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)

c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)

d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)

23 tháng 3 2018

\(\frac{9620}{979}\)

25 tháng 8 2015

dài quá! mình bó tay luôn!

29 tháng 6 2021

a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)

b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)

c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)

29 tháng 6 2021

a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)

b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)

\(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)

c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)

10 tháng 4 2017

\(-\frac{4}{9}\)

16 tháng 7 2016

a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b)\(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)

\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)

\(=\frac{3}{7}.\left(-14\right)\)

\(=-6\)

c)\(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)

\(=9.\left(-\frac{1}{3}\right)^3+9.\frac{1}{27}\)

\(=9.\left[\left(-\frac{1}{3}\right)^3+\frac{1}{27}\right]\)

\(=9.0=0\)

d)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)

\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)

\(=\left(-10\right):\left(-\frac{5}{7}\right)\)

\(=14\)