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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
\(a,-35.12-8.35\)
\(=\left(-35\right)\left(12+8\right)\)
\(=-700\)
\(b,-23.7-7.59-18.7\)
\(=7\left(-23-59-18\right)\)
\(7.\left(-100\right)=-700\)
\(b,115-\left(-12+29\right)\)
\(=115+12-29\)
\(=98\)
\(c,3.\left(-7\right)^2+5.\left(-4\right)^3\)
\(=3.49-5.64\)
\(=147-320\)
\(=-173\)
a)-35.12-8-35=-35(12+8)=-35.20=-700
b)23.(-7)-7.5+(-18).7=-7(23+59+18)=-7.100=-700
c)115-(-12+29)=115+12-29=98
d)\(3\left(-7\right)^2+5\left(-4\right)^3=3.7^2-5.4^3=3.49-5.64=147-320-173\)
a, (-20) + 16 + (-34) + 20 = [(-20) + 20] + [16 + (-34)] = 0 + (-18) = (-18) b, 12 + 3.[39 - (5 - 2)2] = 12 + 3.[39 - 32] = 12 + 3.[39 - 9] = 12 + 3.30 = 12 + 90 = 102
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
\(a,=\dfrac{4}{12}+\dfrac{15}{12}-\dfrac{7}{12}=\dfrac{12}{12}=1\\ b,=1,6\left(2\dfrac{1}{4}-3\dfrac{1}{2}\right)\\ =1,6\cdot\dfrac{-3}{4}=-2,4\)