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a, (\(\dfrac{9}{4}\))5 : (\(\dfrac{1}{4}\))5
= (\(\dfrac{9}{4}\) : \(\dfrac{1}{4}\))5
= 95
= 59049
b, 182 : 92
= (18:9)2
= 22
= 4
c, [(-2)\(^4\)]3
= 212
= 4096
d, 57.(\(\dfrac{1}{5}\))7
= (5.\(\dfrac{1}{5}\))7
= 17
= 1
e, (6,5)3: (6,5)2
= 6,5
\(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{127}-\dfrac{7}{18}+\dfrac{4}{35}-\left(-\dfrac{2}{7}\right)\)
\(=\left[-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right]+\left[\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right]+\dfrac{1}{127}\)
\(=-\dfrac{18}{18}+\dfrac{35}{35}+\dfrac{1}{127}\)
\(=-1+1+\dfrac{1}{127}\)
\(=\dfrac{1}{127}\)
\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}-\left(-\frac{2}{7}\right)=-\frac{1}{2}+\frac{3}{5}-\frac{1}{9}+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}+\frac{2}{7}=\frac{1}{127}\)
= 11/125 - [(17/18 - 4/9) + (5/7 - 17/14)]
= 11/125 - (1/2 - 1/2)
= 11/125
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)=\dfrac{11}{125}-\dfrac{17-4.2}{18}+\dfrac{17-5.2}{14}=\dfrac{11}{125}-\dfrac{9}{18}+\dfrac{7}{14}=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)
a: \(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
b: \(=-12+\dfrac{8}{9}-\dfrac{5}{18}=\dfrac{-216}{18}+\dfrac{16}{18}-\dfrac{5}{18}=\dfrac{-205}{18}\)
c: \(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}=6-\dfrac{19}{7}=\dfrac{23}{7}\)
d: \(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=\dfrac{-1}{4}\cdot20=-5\)
11/125 - 17/18 - 5/7 + 4/9 + 17/14
=11/125 - (17/18 - 4/9) - (5/7 -17/14)
=11/125 - (17/18 - 8/18) - (10/14 - 17/14)
=11/125 - 9/18 + 7/14
=11/125 - 1/2 + 1/2
=11/125 (= 0,88)
- 5 9 : - 7 18 = - 5 9 . - 18 7 = 10 7