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a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)
\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)
\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)
b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)
\(=-\frac{5}{72}:\frac{3.7}{8.9}\)
\(=-\frac{5}{72}:\frac{7}{24}\)
\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)
a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)
b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)
c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)
\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)
\(=\frac{-23}{25}+\frac{-27}{25}\)
\(=\frac{-50}{25}=-2\)
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
`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
a) Ta có: \(-19-\left|+13\right|+\left|-10\right|-\left(-5\right)-11\)
\(=-19-13+10+5-11\)
\(=-28\)
b) Ta có: \(\left(-10\right)-\left(11\right)+\left|-5\right|-\left|-7\right|-8\)
\(=-10-11+5-7-8\)
\(=-31\)
c) Ta có: \(-\left(-8\right)-\left(+7\right)-\left|-11\right|+\left|+12\right|\)
\(=8-7-11+12\)
\(=2\)
d) Ta có: \(-14-\left|-18\right|-\left(-20\right)-\left|-25\right|\)
\(=-14-18+20-25\)
\(=-37\)
đ) Ta có: \(23-\left|-15\right|-\left(-17\right)+\left|-13\right|\)
\(=23-15+17+13\)
\(=38\)
a: =-5/9-4/9+8/15+7/15-2/11=-2/11
b: =10/17+7/17-5/13-8/13+11/25
=11/25
c: =(9/12-2/12)*3/2=7/12*3/2=21/24=7/8
d: =(31/10-25/10)*3-2
=3/5*3-2
=9/5-2
=-1/5
a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)
b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)
c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)
\(a)\)Ta tách thành 2 vế của phép tính ra thành:
\(\left(5+7+9+11+13+15+17\right)+\left(3+8+13+18+23+28\right)\)
Ta gọi dãy \(\left(5+7+9+11+13+15+17\right)\)là \(S_1\)
\(\left(3+8+13+18+23+28\right)\) là dãy \(S_2\)
- Số số hạng của dãy \(S_1\)là:
\(\left(17-5\right)\div2+1=7\)( số hạng )
Tổng của dãy \(S_1\)là:
\(\left(17+5\right)\times7\div2=77\)
- Số số hạng của dãy \(S_2\)là:\(\left(3+8+13+18+23+28\right)\)
\(\left(28-3\right)\div5+1=6\)( số hạng )
Tổng của dãy \(S_2\)là:
\(\left(28+3\right)\times6\div2=93\)
Tổng của dãy \(S_1\)và \(S_2\)là:
\(77+93=170\)
Đáp số: \(170\)
\(b)\)Ta ghép thành 2 vế của phép tính ra thành:
\(\left(4+7+10+13+16+19\right)+\left(5+9+13+17+21+25\right)\)
Ta gọi dãy \(\left(4+7+10+13+16+19\right)\)là \(S_1\)
\(\left(5+9+13+17+21+25\right)\)là \(S_2\)
- Số số hạng của dãy số \(S_1\)là
\(\left(19-4\right)\div3+1=6\)( số hạng )
Tổng của dãy số \(S_1\)là:
\(\left(19+4\right)\times6\div2=69\)
- Số số hạng của dãy \(S_2\)là:\(\left(5+9+13+17+21+25\right)\)
\(\left(25-5\right)\div4+1=6\)( số hạng )
Tổng của dãy số \(S_2\)là:
\(\left(25+5\right)\times6\div2=90\)
Tổng của dãy \(S_1\) và \(S_2\)là:
\(69+90=159\)
Đáp số: \(159\)
a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
Đáp án cần chọn là: B
A = 10 17 − 5 13 − − 7 17 − 8 13 + 11 25 = 10 17 − − 7 17 + − 5 13 − 8 13 + 11 25 = 1 − 1 + 11 25 = 11 25