tính
a. 1/2 . ( 4/3 + 2/5 ) - 3/4 . ( 8/9 + 13/3 )
b. 1/5 : 1/10 - 1/3 . ( 6/5 - 9/4 )
c. -3/4 . ( 20/9 - 8/15 ) - 5/3 . I -9/10 I
p/s: mình đang cần gấp, bạn nào giúp mình thì mình sẽ tick cho nhé
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7 tháng 7 2021
a)=1/2 . 8/15 - 3/4.47/9
=4/15 - 47/12
=-73/20
b)=2-1/3 . -21/20
=2+7/20
=47/20
1 tháng 5 2019
a, 13/6+5/8 : -3/4 - 7/12.4
= 13/6 + -5/6-7/3
=8/6-7/3
= -6/6
= -1
b, ( 73/5 - 21/3) + ( 4/3-43/5 )
= 73/5-21/3+4/3-43/5
=( 73/5-43/5)-(21/3-4/3)
= 6-17/3
=1/3
c, 7/5.4/9 +7/5: 9/16- 14/10.2/9
= 7/5.4/9 +7/5.16/9 - 14/45
=7/5.(4/9+16/9)-14/45
=7/5.20/9-14/45
= 140/45 - 14/45
= 126/45
Xong rùi nè! Nhưng bạn kiểm tra lại giùm nhé vì làm vào ban đêm nên hơi bất tiện
11 tháng 7 2017
Tính
a) \(\dfrac{1}{2}\) . ( \(\dfrac{4}{3}\) + \(\dfrac{2}{5}\) ) - \(\dfrac{3}{4}\) . ( \(\dfrac{8}{9}\) + \(\dfrac{13}{3}\) )
= \(\dfrac{1}{2}\) . \(\dfrac{8}{15}\) - \(\dfrac{3}{4}\) . \(\dfrac{47}{9}\)
= \(\dfrac{4}{15}\) - \(\dfrac{47}{12}\)
= \(\dfrac{-73}{20}\)
b) \(\dfrac{1}{5}\) : \(\dfrac{1}{10}\) - \(\dfrac{1}{3}\) . ( \(\dfrac{6}{5}\)-\(\dfrac{9}{4}\) )
= 2 - \(\dfrac{1}{3}\) . \(\dfrac{-21}{20}\)
= 2 - \(\dfrac{-7}{20}\)
= \(\dfrac{47}{20}\)
c) \(\dfrac{-3}{4}\) . ( \(\dfrac{20}{9}\) - \(\dfrac{8}{15}\) ) - \(\dfrac{5}{3}\) . \(\dfrac{9}{10}\)
= \(\dfrac{-3}{4}\) . \(\dfrac{76}{45}\) - \(\dfrac{3}{2}\)
= \(\dfrac{-19}{15}\) - \(\dfrac{3}{2}\)
= \(\dfrac{-7}{30}\)
NN
1 tháng 8 2020
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
NN
1 tháng 8 2020
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
6 tháng 8 2019
a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)
=\(9^4.13:9^3=13.9=117\)
b) 100-(75-25)=100-50=50
14 tháng 7 2016
B=\(1+3^2+3^4+...+3^{100}\)
9B=\(3^2+3^4+...+3^{100}\)
9B-B=\(\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
8B=\(3^{102}-1\)
B=\(\left(3^{102}-1\right):8\)
14 tháng 7 2016
C=\(1+5^3+5^6+...+5^{99}\)
125C=\(5^3+5^6+5^9+...+5^{102}\)
125C-C=\(\left(5^3+5^6+5^9+...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
124C=\(5^{102}-1\)
C=\(\left(5^{102}-1\right):124\)
