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

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

Trời ạ. 7 3/5 chứ k phải 73/5 -_-

ghi rõ thêm từ và nhé!

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) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)

\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)

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) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)

bạn juventus ơi , bạn làm nốt cho mình mấy câu sau với 

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) \(\dfrac{-3}{4}\) . ( \(\dfrac{20}{9}\) - \(\dfrac{8}{15}\) ) - \(\dfrac{5}{3}\) . \(\dfrac{9}{10}\)

= \(\dfrac{-3}{4}\) . \(\dfrac{76}{45}\) - \(\dfrac{3}{2}\)

= \(\dfrac{-19}{15}\) - \(\dfrac{3}{2}\)

= \(\dfrac{-7}{30}\)

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=\(\left(3^{102}-1\right):8\)

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=\(\left(5^{102}-1\right):124\)

=3 nha bn

KANZAKI MIZUKI GIẢI TRINH TỰ HỘ MÌNH VỚI

BẠN GIỎI TOÁN NHƯ VẬY GIÚP MÌNH VỚI

địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(\dfrac{3}{5}\)+4-\(\dfrac{6}{7}\)=\(\dfrac{3}{5}\)+\(\dfrac{20}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{23}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{161}{35}\)-\(\dfrac{30}{35}\)=\(\dfrac{131}{35}\)