1, \(\left(\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}\right)-\left(\dfrac{15}{33}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)
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\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{15}\right)+\dfrac{-6}{33}\right]+\dfrac{-3}{4}\)
\(=\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{5}{11}\cdot\dfrac{23}{13}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{115}{143}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\dfrac{-141}{143}-\dfrac{3}{4}\)
\(=-\dfrac{141}{104}-\dfrac{3}{4}\)
\(=-\dfrac{219}{104}\)
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: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
a) \(\dfrac{3}{5}:\left(\dfrac{-1}{15}-\dfrac{1}{6}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-1\dfrac{1}{15}\right)\)
\(=\dfrac{3}{5}:\left(\dfrac{-1}{15}-1\dfrac{1}{15}-\dfrac{1}{6}+\dfrac{-2}{6}\right)\)
\(=\dfrac{3}{5}:\left(\dfrac{-17}{15}-\dfrac{1}{6}\right)\)
\(=\dfrac{3}{5}\cdot\dfrac{-10}{13}\)
\(=\dfrac{-6}{13}\)
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
\(=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{48}{49}\)
\(=1+\dfrac{7}{33}-\dfrac{18}{33}-\dfrac{5}{3}+\dfrac{48}{49}\)
\(=\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{48}{49}=\dfrac{48}{49}-1=-\dfrac{1}{49}\)
Dấu ''.'' ở gần số 1 đúng ko bạn