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a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)
b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)
c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)
\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a, \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{-3}{8}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{25}{144}+\dfrac{1}{2}\)
\(=\dfrac{97}{144}\)
b, \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)
\(=\dfrac{28}{15}.0,75-\dfrac{4}{5}:\dfrac{7}{5}\)
\(=\dfrac{7}{5}-\dfrac{4}{7}\)
\(=\dfrac{29}{35}\)
a)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
\(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8})}{5\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}\)
\(=1\)