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Phương Anh Ribi
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Tóc Em Rối Rồi Kìa
26 tháng 3 2018 lúc 19:15

\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)

\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)

\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)

\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)

\(2B=3B-B\)

\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)

\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)

\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)

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Đinh Quốc Vĩ
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Anh Tuấn Đào
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Nguyễn Lê Phước Thịnh
2 tháng 7 2023 lúc 8:31

a: =-21/36-3/36=-24/36=-2/3

b: =43/12*1/2+5/24=43/24+5/24=2

c: =8/9+1/9=1

e: =1-1/4+1/4-1/7+...+1/97-1/100

=1-1/100=99/100

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Nguyễn Quý Vương
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Nguyễn Quang Đạt
21 tháng 6 2022 lúc 16:43

A=14+112+136+...+1972+12916

3A=34+14+112+...+1324+1972

3A−A=(34+14+112+...+1324+1972)−(14+112+136+...+1972+12916)

2A=34−12916

A=10932916

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Minh Ngọc
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Minh Lệ
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Gia Linh
24 tháng 8 2023 lúc 1:21

a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)

b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)

c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)

d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)

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Chử Hải Yến
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Nguyễn Lê Phước Thịnh
29 tháng 5 2022 lúc 19:36

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

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Trần Thu Hiền
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kimcherry
2 tháng 5 2022 lúc 15:47

a. 5/6

b. -6/7

c. 2/3

d. 2/9

e. -63/325

g. 43/20

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kimcherry
2 tháng 5 2022 lúc 16:16

a. 1/12 + 3/4

= 1/12+ 9/12

= 10/12 = 5/6

b. -4/7. 1 1/2

= -4/7. 3/2

= -6/7

c. 7/9+ ( 2/3 + -7/9)

= 7/9 + -1/9

= 8/9 = 2/3

d. 2/3 - 1/3 : 3/4

= 2/3 - 4/9

= 6/9 - 4/9

= 2/9

e. -7/25 . 11/13+ - 7/25. 2/13

=-77/ 325+ -14/ 325

= - 7/25

g. 2 2/5 . 0,25- ( 11/20+ 75%) : 13/5

= 12/5 . 1/4 - ( 11/20 + 3/4) : 13/5

= 12/5 . 1/4 - 13/10 : 12/ 5

= 3/5 - 13/24

= 7/120

lm đại vậy thôi nha hơi lâu vì mik bị lag nên phải làm lại

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nguyễn trần minh
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Quỳnh Như
16 tháng 9 2017 lúc 9:00

a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=1-1+\dfrac{1}{72}\)

\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)

\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)

\(=-\left(-\dfrac{173}{1287}\right)\)

\(=\dfrac{173}{1287}\)

c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{-49}{50}\)

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