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9 tháng 8 2016

    \(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{9}.\frac{1}{10}\)

\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{10}\)

\(=\frac{9}{10}\)

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

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

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

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

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

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

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

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

7 tháng 8 2020

a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79

= 1/10 - 1/79

= máy tính ok

mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha

7 tháng 8 2020

a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)

26 tháng 6 2018

\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)

\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)

\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)

\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)

\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)

Hok tốt

25 tháng 10 2024

3 phần 5 + 7 phần 10

5 tháng 8 2020

Dấu này * là dấu nhân

Một năm rồi không có ai trả lời à 

7 tháng 3 2018

1,

a,-3/5

b,-1/2

c,19/39

d,1/4

e,-39/40

f,-59/56

2,

a,=

b,<

c,>

d,<

k cho mình nha

15 tháng 4 2020

a, \(\frac{3}{5}-\left(-\frac{7}{10}\right)-\frac{13}{20}=\frac{3}{5}+\frac{7}{10}-\frac{13}{20}=\frac{12}{20}+\frac{14}{20}-\frac{13}{20}=\frac{13}{20}\)

b, \(\frac{3}{4}+\frac{-1}{3}-\frac{5}{18}=\frac{27}{36}+\frac{-12}{36}-\frac{10}{36}=\frac{5}{36}\)

c, \(\frac{3}{14}-\frac{5}{-8}+\frac{-1}{2}=\frac{12}{56}+\frac{35}{56}+\frac{-28}{56}=\frac{19}{56}\)

d, \(\frac{1}{2}+\frac{1}{-3}+\frac{1}{4}-\left(-\frac{1}{6}\right)=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6}{12}+\frac{-4}{12}+\frac{3}{12}+\frac{2}{12}=\frac{7}{12}\)

đ, \(\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}=\frac{19}{24}-\frac{12}{24}-\frac{8}{24}-\frac{7}{24}=-\frac{8}{24}=-\frac{1}{3}\)

g, \(-\frac{1}{3}+\frac{1}{2}-\frac{5}{6}=-\frac{2}{6}+\frac{3}{6}-\frac{5}{6}=-\frac{4}{6}=\frac{-2}{3}\)

e, \(\frac{4}{-9}+\frac{8}{5}+\frac{-5}{9}-\frac{2}{11}+\frac{7}{15}=\frac{-220}{495}+\frac{792}{495}+\frac{-275}{495}-\frac{90}{495}+\frac{231}{495}=\frac{438}{495}=\frac{146}{165}\)

ê, \(\frac{7}{11}-\left(\frac{3}{5}+\frac{7}{11}\right)=\frac{7}{11}-\frac{3}{5}-\frac{7}{11}=\left(\frac{7}{11}-\frac{7}{11}\right)-\frac{3}{5}=0-\frac{3}{5}=-\frac{3}{5}\)

Dài quá không nhanh được @@

15 tháng 4 2020

Cảm ơn bạn nhiều nha.

6 tháng 8 2019

Mọi người ơi, giúp em với ạ

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