Tính nhanh:
A=\(\left(\frac{2}{5}+\frac{2}{9}+\frac{2}{11}\div\frac{7}{5}+\frac{7}{9}+\frac{7}{11}\right)\div\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\div\frac{7}{6}-\frac{7}{8}+\frac{7}{10}\right)\)
K
Khách
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Những câu hỏi liên quan
25 tháng 3 2017
a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7
= -3/7. ( 5/11 + 6/11 ) + 9/7
= -3/7. 1 + 9/7
= -3/7 + 9/7
= 6/7
b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4
= 13/13 + -5/5 - 3/4
= 1 + (-1) - 3/4
= 0 - 3/4
= -3/4
c) = -19/17. 4/7 + 19/17. -3/7 + 19/17
= 19/17. -4/7 + 19/17. -3/7 + 19/17.1
= 19/17.( -4/7 + -3/7 + 19/17
= 19/17. -7/7 + 19/17
= 19/17. (-1) + 19/17
= -19/17 + 19/17
= 0
tk mk nha,thanks
KK
29 tháng 7 2019
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
29 tháng 7 2019
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
17 tháng 2 2022
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
EG
17 tháng 8 2018
\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)
KN
14 tháng 5 2019
\(9\frac{2}{9}+\frac{2}{3}+7\frac{7}{9}\)
\(=9+\frac{2}{9}+\frac{2}{3}+7+\frac{7}{9}\)
\(=\left(9+7\right)+\left(\frac{2}{9}+\frac{7}{9}\right)+\frac{2}{3}\)
\(=16+1+\frac{2}{3}\)
\(=17+\frac{2}{3}\)
\(=\frac{51}{3}+\frac{2}{3}\)
\(=\frac{53}{3}\)
KN
14 tháng 5 2019
\(\frac{5}{9}.\frac{10}{11}+\frac{5}{9}.\frac{14}{11}-\frac{5}{9}.\frac{15}{11}\)
\(=\frac{5}{9}\left(\frac{10}{11}+\frac{14}{11}-\frac{15}{11}\right)\)
\(=\frac{5}{9}.\frac{9}{11}\)
\(=\frac{5}{11}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)
b)
\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)
d)
\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}\)
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)