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bài1
a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\)
=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\)
=\(\dfrac{1}{12}+\dfrac{9}{12}\)
=\(\dfrac{10}{12}=\dfrac{5}{6}\)
bài 1
b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\)
= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\)
= \(-\dfrac{6}{5}+\dfrac{3}{10}\)
=\(-\dfrac{12}{10}+\dfrac{3}{10}\)
=\(-\dfrac{9}{10}\)
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
\(a,\dfrac{15^3}{5^4}\)
\(=\dfrac{\left(3\cdot5\right)^3}{5^4}\)
\(=\dfrac{3^3\cdot5^3}{5^4}\)
\(=\dfrac{3^3}{5}\)
\(=\dfrac{27}{5}\)
\(---\)
\(b,\dfrac{21^3}{7^4}\)
\(=\dfrac{\left(3\cdot7\right)^3}{7^4}\)
\(=\dfrac{3^3\cdot7^3}{7^4}\)
\(=\dfrac{3^3}{7}\)
\(=\dfrac{27}{7}\)
\(---\)
\(c,\dfrac{6^6}{3^8}\)
\(=\dfrac{\left(2\cdot3\right)^6}{3^8}\)
\(=\dfrac{2^6\cdot3^6}{3^8}\)
\(=\dfrac{2^6}{3^2}\)
\(=\dfrac{64}{9}\)
#\(Toru\)
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
a: \(=\dfrac{3}{8}\left(72+\dfrac{1}{5}-51-\dfrac{1}{5}\right)=\dfrac{3}{8}\cdot21=\dfrac{63}{8}\)
b: \(=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{2}-\dfrac{1}{2}=-1\)
c: \(=4\left(35+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}-\left(45+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}\)
\(=\dfrac{-1}{5}\left(140+\dfrac{2}{3}-45-\dfrac{1}{6}\right)=-\dfrac{191}{10}\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
a: \(A=\left[6\cdot\dfrac{1}{27}+3\cdot\dfrac{1}{3}+1\right]:\dfrac{-4}{3}\)
\(=\left(\dfrac{2}{9}+2\right)\cdot\dfrac{-3}{4}\)
\(=\dfrac{20}{9}\cdot\dfrac{-3}{4}=\dfrac{-60}{36}=\dfrac{-5}{3}\)
b: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)