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a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)
b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
\(=\dfrac{17}{8}:\dfrac{51}{40}-\left(15-13-\dfrac{1}{3}\right):\dfrac{25}{6}\)
\(=\dfrac{5}{3}-\dfrac{5}{3}\cdot\dfrac{6}{25}=\dfrac{5}{3}\cdot\dfrac{19}{25}=\dfrac{19}{15}\)
\(2\dfrac{1}{8}:1\dfrac{11}{40}-\left(15-13\dfrac{1}{3}\right):4\dfrac{1}{6}\)
\(=\dfrac{17}{8}.\dfrac{40}{51}-\left(15-\dfrac{40}{3}\right).\dfrac{6}{25}\)
\(=\dfrac{17}{8}.\dfrac{40}{51}-\dfrac{5}{3}.\dfrac{6}{25}\)
\(=\dfrac{5}{3}-\dfrac{2}{5}\)
\(=\dfrac{19}{15}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
\(\dfrac{5}{2x1}+\dfrac{4}{1x11}+\dfrac{3}{11x2}+\dfrac{1}{2x15}+\dfrac{13}{15x4}+\dfrac{15}{4x13}\)
=7x(\(\dfrac{5}{2x7}+\dfrac{4}{7x11}+\dfrac{3}{11x14}+\dfrac{1}{14x15}+\dfrac{13}{15x28}+\dfrac{15}{28x43}\))
=7x\(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{43}\)=7x(\(\dfrac{1}{2}-\dfrac{1}{43}\))
=7x\(\dfrac{41}{86}\)
=\(\dfrac{287}{86}\)
5/2x1+4/1x11+3/11x2+1/2x15+13/15x4+15/4x43=7x(5/2x7+4/7x11+3/11x14+1/14x15+13/15x28+15/28x43)=7x(1/2-1/7+1/7-1/11+1/11-1/14+1/14+1/15+1/15-1/28+1/28-1/43)=7x(1/2-1/43)=7x41/86=287/86
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
-11/12
-11/12