-34% : \(\dfrac{51}{16}\) - \(3\dfrac{7}{9}\) . 6,5 + (0,4)2
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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
\(A=\dfrac{15}{34}+\dfrac{27}{21}+\dfrac{9}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)
\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1\dfrac{15}{17}\right)+\left(\dfrac{27}{21}+\dfrac{2}{3}\right)\)
\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-\dfrac{32}{17}\right)+\left(\dfrac{27}{21}+\dfrac{2}{3}\right)\)
\(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-\dfrac{64}{34}\right)+\left(\dfrac{27}{21}+\dfrac{14}{21}\right)\)
\(A=\dfrac{-20}{17}+\dfrac{41}{21}\)
\(A=\dfrac{-420}{357}+\dfrac{697}{357}=\dfrac{277}{357}\)
\(B=16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)-28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
\(B=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)-28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)\)
\(B=\left(-\dfrac{5}{3}\right)\left(16\dfrac{2}{7}-28\dfrac{2}{7}\right)\)
\(B=\left(-\dfrac{5}{3}\right)\left(-12\right)\)
\(B=20\)
a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)
\(=-\dfrac{4}{3}\cdot0.4\)
\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)
b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{7}\)
\(=\dfrac{6}{7}\)
a: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)
b: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\dfrac{3}{4}\)
\(\dfrac{\left(17\dfrac{8}{19}-16\dfrac{9}{18}\right).\left(17,5+16\dfrac{17}{51}-32\dfrac{15}{22}\right)}{\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}}\)
=\(\dfrac{\dfrac{35}{38}.\dfrac{38}{33}}{\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\left(\dfrac{1}{3}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\dfrac{10}{33}}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{33}}\)
=\(\dfrac{35}{33}:\dfrac{7}{33}\)
=\(\dfrac{35}{33}.\dfrac{33}{7}\)
=5