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\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)
2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)
\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)
3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)
a: \(=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{48}{49}\)
\(=\dfrac{-3}{2}+\dfrac{7}{33}-\dfrac{18}{33}-\dfrac{5}{3}+\dfrac{48}{49}\)
\(=\dfrac{-9-10}{6}+\dfrac{-11}{33}+\dfrac{48}{49}\)
\(=\dfrac{-19}{6}+\dfrac{-1}{3}+\dfrac{48}{49}=-\dfrac{247}{98}\)
b: \(=\dfrac{11}{125}-\dfrac{17}{18}+\dfrac{4}{9}-\dfrac{5}{7}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\dfrac{-17+8}{9}+\dfrac{-10+17}{14}\)
\(=\dfrac{11}{125}-1+\dfrac{7}{14}=\dfrac{11}{125}+\dfrac{1}{2}=\dfrac{22+125}{250}=\dfrac{147}{250}\)
a) \(\frac{3}{4}+\left(\frac{-1}{3}\right)-\frac{5}{18}=\frac{5}{36}\)
b) \(-2+\left(\frac{-5}{8}\right)+\frac{3}{4}=\frac{-15}{8}\)
c) \(\frac{15}{8}.\frac{12}{5}-\frac{11}{3}=\frac{5}{6}\)
d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}=\frac{1}{9}\)
mk chỉ đưa ra kp thôi!
a) 5/9 + 4/9 . 3/7 + 4/9 . 4/7
= 5/9 + 4/9 . (3/7 + 4/7)
= 5/9 + 4/9 . 1
= 5/9 + 4/9
= 1
=\(-\frac{1}{4}+\frac{7}{33}-\frac{5}{3}+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}=\left(-\frac{1}{4}-\frac{5}{3}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=\left(\frac{-3}{12}-\frac{20}{12}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{18}{33}\right)+\frac{48}{49}\)
\(=-\frac{8}{12}+\frac{-11}{33}+\frac{48}{49}=-\frac{2}{3}-\frac{1}{3}+\frac{48}{49}=-\frac{3}{3}+\frac{48}{49}=-1+\frac{48}{49}=-\frac{49}{49}+\frac{48}{49}=-\frac{1}{49}\)