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a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
34%:51/16-3 = -217/75
7/9x6,5-(0,4) mũ 2 = 2203/450
suy ra 2203/450 lớn hơn lúc này ta có 7/9x6,5-(0,4)mũ 2 lớn hơn 34%:51/16-3
34%:51/16-3 = -217/75
7/9x6,5-(0,4) mũ 2 = 2203/450
suy ra 2203/450 lớn hơn lúc này ta có 7/9x6,5-(0,4)mũ 2 lớn hơn 34%:51/16-3
\(\dfrac{\left(17\dfrac{8}{19}-16\dfrac{9}{18}\right).\left(17,5+16\dfrac{17}{51}-32\dfrac{15}{22}\right)}{\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}}\)
=\(\dfrac{\dfrac{35}{38}.\dfrac{38}{33}}{\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\left(\dfrac{1}{3}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\dfrac{10}{33}}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{33}}\)
=\(\dfrac{35}{33}:\dfrac{7}{33}\)
=\(\dfrac{35}{33}.\dfrac{33}{7}\)
=5
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
\(B=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left[\dfrac{2\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{1\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}{\dfrac{7}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right]\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{2}{7}-1:\dfrac{7}{2}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=0\cdot\dfrac{1842009}{1842010}=0\)