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7 tháng 8 2020
\(a,\frac{-1}{9}.\frac{15}{22}.\frac{-9}{25}\)
\(=\frac{-1.15.\left(-9\right)}{9.22.25}\)
\(=\frac{3}{110}\)
\(b,\frac{-2}{7}.\left(\frac{5}{13}-\frac{9}{15}\right)-\frac{2}{7}.\frac{8}{13}\)
\(=\frac{-2}{7}.\left(\frac{5}{13}+\frac{8}{13}-\frac{3}{5}\right)\)
\(=\frac{-2}{7}.\left(1-\frac{3}{5}\right)\)
\(=\frac{-2}{7}.\frac{2}{5}\)
\(=\frac{-4}{35}\)
\(c,\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}\right)-\frac{3}{10}.\left(\frac{5}{9}-\frac{3}{5}\right)\)
\(=\frac{3}{10}.\left[\left(\frac{-4}{9}+\frac{2}{5}\right)-\left(\frac{5}{9}-\frac{3}{5}\right)\right]\)
\(=\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}-\frac{5}{9}+\frac{3}{5}\right)\)
\(=\frac{3}{10}.\left[\left(\frac{-4}{9}-\frac{5}{9}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{3}{10}.\left(-1+1\right)\)
\(=\frac{3}{10}.0\)
\(=0\)
\(d,\frac{4}{11}-\frac{5}{13}+\frac{7}{11}-\frac{8}{13}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)+\left(\frac{-5}{13}-\frac{8}{13}\right)\)
\(=1-1\)
\(=0\)
Học tốt
29 tháng 4 2017
a, \(\left(4\dfrac{1}{9}+3\dfrac{1}{4}\right).2\dfrac{1}{4}+2\dfrac{3}{4}\)
\(=\left(\dfrac{37}{9}+\dfrac{13}{4}\right).\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265}{36}.\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265}{16}+\dfrac{11}{4}\)
\(=\dfrac{309}{16}\)
b, \(\dfrac{9}{23}.\dfrac{5}{8}+\dfrac{9}{23}.\dfrac{3}{8}-\dfrac{9}{23}\)
\(=\dfrac{45}{184}+\dfrac{27}{184}-\dfrac{9}{23}\)
\(=\dfrac{9}{23}-\dfrac{9}{23}\)
\(=\dfrac{1}{1}\)
c, \(1+\left(\dfrac{9}{10}-\dfrac{4}{5}\right)\div3\dfrac{1}{6}\)
\(=1+\left(\dfrac{9}{10}-\dfrac{4}{5}\right)\div\dfrac{19}{6}\)
\(=1+\dfrac{1}{10}\div\dfrac{19}{6}\)
\(=1+\dfrac{3}{95}\)
\(=1\dfrac{3}{95}\)
d, ???
HT
7 tháng 8 2015
B = \(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
B = \(\frac{2}{10.18}+\frac{2}{18.26}+\frac{2}{26.34}+...+\frac{2}{802.810}\)
B = \(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{8}{81}\)
B = \(\frac{2}{81}\)
AP
2
HP
29 tháng 12 2015
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
21 tháng 4 2017
a) \(\left(4\dfrac{1}{9}+3\dfrac{1}{4}\right).2\dfrac{1}{4}+2\dfrac{3}{4}\)
\(=\left(\dfrac{37}{9}+\dfrac{13}{4}\right).\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(\dfrac{148}{36}+\dfrac{117}{36}\right).\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(\dfrac{148+117}{36}\right).\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265}{36}.\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265.9}{36.4}+\dfrac{11}{4}\)
\(=\dfrac{265.1}{4.4}+\dfrac{11}{4}\)
\(=\dfrac{265}{16}+\dfrac{11}{4}\)
\(=\dfrac{265}{16}+\dfrac{44}{16}\)
\(=\dfrac{309}{16}\)
AG
0
\(a,\left(-\dfrac{13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)\\ =-\dfrac{13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =-\dfrac{3}{7}.\)
a) $(-\frac{13}{7}-\frac49)-(-\frac{10}{7}-\frac49)$
$=-\frac{13}{7}-\frac49+\frac{10}{7}+\frac49$
$=(-\frac{13}{7}+\frac{10}{7})+(-\frac49+\frac49)$
$=\frac{-3}{7}$