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`a)3/5+(-4/9)`
`=3/5-4/9`
`=27/45-20/45`
`=7/45`
`b)3/5+2/5 . 15/8`
`=3/5 + 30/40`
`=3/5+3/4`
`=12/20+15/20`
`=27/20`
`c)7/2 . 8/13 + 8/13 . (-5/2)`
`=8/13 . (7/2 +(-5)/2)`
`=8/13 . 1`
`=8/13`
`d)-5/17 . (-9/23)+9/23 . (-22/17) + 11 9/23`
`=-5/17 . (-9/23) + 9/23 . (-1) . 22/17 + 11 + 9/23`
`=-5/17 . (-9/23) + (-9/23) . 22/17+11+9/23`
`= -9/23 ( -5/17 + 22/17)+11+9/23`
`= - 9/23 . 1+11+9/23`
`=-9/23+11+9/23`
`=(-9/23+9/23)+11`
`=0+11`
`=11`
a: =27/45-20/45=7/45
b: =3/5+30/40
=3/5+3/4
=12/20+15/20
=27/20
c: =8/13(7/2-5/2+1)=8/13*2=16/13
d: =9/23*5/17-9/23*22/17+11+9/23
=-9/23+11+9/23
=11
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
2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
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, \(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+...+\(\dfrac{5}{99.101}\)
= 5.\(\dfrac{1}{1.3}\)+5.\(\dfrac{1}{3.5}\)+5.\(\dfrac{1}{5.7}\)+...+5.\(\dfrac{1}{99.101}\)
=5.(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{99.101}\))
=5.(\(\dfrac{2}{2}\).\(\dfrac{1}{1.3}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{3.5}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{5.7}\)+...+\(\dfrac{2}{2}\).\(\dfrac{1}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).\(\dfrac{100}{101}\)
=\(\dfrac{250}{101}\)
=\(2\dfrac{48}{101}\)
b,\(\dfrac{-11}{23}\).\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\).\(\dfrac{-11}{23}\)-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).(\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\))-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).2-\(\dfrac{1}{23}\)
=\(\dfrac{-22}{23}\)-\(\dfrac{1}{23}\)
=-1
c,\(\dfrac{2.3}{7}\)+(\(\dfrac{2}{9}\)-\(1\dfrac{1}{3}\))-\(\dfrac{5}{3}\):\(\dfrac{1}{9}\)
=\(\dfrac{6}{7}\)+(\(\dfrac{2}{9}\)-\(\dfrac{4}{3}\))-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-15
=\(\dfrac{54}{63}\)-\(\dfrac{70}{63}\)-\(\dfrac{945}{63}\)
=\(\dfrac{-961}{63}\)=\(-15\dfrac{16}{63}\)
d,(20+\(9\dfrac{1}{4}\)):\(2\dfrac{1}{4}\)
=(20+\(\dfrac{37}{4}\)):\(\dfrac{9}{4}\)
=20:\(\dfrac{9}{4}\)+\(\dfrac{37}{4}\):\(\dfrac{9}{4}\)
=20.\(\dfrac{4}{9}\)+\(\dfrac{37}{4}\).\(\dfrac{4}{9}\)
=\(\dfrac{80}{9}\)+\(\dfrac{37}{9}\)
=\(\dfrac{117}{9}\)
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