Tính nhanh: a) A= 1 \(\dfrac{1}{5} \dfrac{1}{25} \dfrac{1}{125} \dfrac{1}{625} ... \dfrac{1}{78125}\) b) B= \(\dfrac{1}{3} \dfrac{1}{12} \dfrac{1}{48} \dfrac{1}{192} \dfrac{1}{768} ... \dfrac{1}{3...

NN

Ngô Ngọc Khánh

11 tháng 3 2018

Tính nhanh: a) A= 1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\) b) B= \(\dfrac{1}{3}+\dfrac{1}{12}+\dfrac{1}{48}+\dfrac{1}{192}+\dfrac{1}{768}+...+\dfrac{1}{36864}\) c) M= \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9900}\) d) P=...

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#Toán lớp 6

0

PD

Phạm Đàm Tuệ Đăng

23 tháng 5 2022 - olm

1,thực hiện phép tính bằng cách thuận tiện nhất

a,1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)

#Toán lớp 5

1

HC

Hei Cheng

23 tháng 5 2022

\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)

\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)

\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)

\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)

\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)

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DT

Dinh Thi Ngoc Huyen

13 tháng 9 2017

Thuc hien phep tinh; a/ \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\) b/ \(\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\) c/ \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)+ \(\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\) d/...

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 5 2022

a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)

b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)

\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)

c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

=1/4+3/4

=1

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DD

Đỗ Diệu Linh

13 tháng 9 2017

tinh

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

#Toán lớp 7

2

NT

Nguyễn Thanh Hằng

13 tháng 9 2017

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\)

\(=1\)

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NT

Nguyễn Thị Hồng Nhung

13 tháng 9 2017

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)

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T

títtt

9 tháng 10 2023

1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)

2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)

#Toán lớp 11

1

NL

Nguyễn Lê Phước Thịnh

9 tháng 10 2023

1: \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^9}\)

\(=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^9\)

u1=1; q=1/3

\(S_9=\dfrac{u1\cdot\left(1-q^9\right)}{1-q}=\dfrac{1\left(1-\left(\dfrac{1}{3}\right)^9\right)}{1-\dfrac{1}{3}}\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3^9}\right)\)

2:

\(S=\left(\dfrac{1}{5}\right)^0+\left(\dfrac{1}{5}\right)^1+...+\left(\dfrac{1}{5}\right)^7\)

\(u1=1;q=\dfrac{1}{5}\)

\(S_7=\dfrac{1\cdot\left(1-q^7\right)}{1-q}=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}=\dfrac{5}{4}\left(1-\dfrac{1}{5^7}\right)\)

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T

títtt

1 tháng 10 2023

1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)

2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)

#Toán lớp 11

1

NH

Nguyễn Huy Tú

2 tháng 10 2023

1, Ta có \(\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3};\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3};...\)

-> Là cấp số nhân, q = 1/3

Ta có \(S_9=1.\dfrac{1-\left(\dfrac{1}{3}\right)^9}{1-\left(\dfrac{1}{3}\right)}\approx1,5\)

b, Ta có \(\dfrac{\dfrac{1}{5}}{1}=\dfrac{1}{5};\dfrac{\dfrac{1}{25}}{\dfrac{1}{5}}=\dfrac{1}{5};...\)

-> Là cấp số nhân, q = 1/5

\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)

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AM

Anime Miku Cherry Mizuki Lovely

27 tháng 7 2017

Bài 1 : Tính a) A=\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)\(+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\) b)B=\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\) c)C=\(\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\) 1 tiếng nữa mình cần rồi giúp mình nhé...

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#Toán lớp 6

0

GD

gấu đáng yêu **

30 tháng 6 2017

Tính :

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{4}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

#Toán lớp 7

1

AT

An Trịnh Hữu

30 tháng 6 2017

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)

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BM

bé mèo miu

1 tháng 10 2023

a) 0,25-\(\dfrac{2}{3}\)+1\(\dfrac{1}{4}\)b) \(\dfrac{3^2}{2}\):\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\).2010c) {[(\(\dfrac{1}{25}\)-0,6)2:\(\dfrac{49}{125}\)].\(\dfrac{5}{6}\)}-[(\(\dfrac{-1}{3}\))+\(\dfrac{1}{2}\)]d) (-\(\dfrac{1}{2}\)-\(\dfrac{1^{ }}{3}\))2:[(\(\dfrac{-5}{36}\))-(\(\dfrac{-5}{36}\))0]Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like...

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#Toán lớp 6

1

T

Toru

1 tháng 10 2023

a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)

\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)

\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)

\(=\dfrac{10}{12}\)

\(=\dfrac{5}{6}\)

\(---\)

b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)

\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)

\(=18+\dfrac{3015}{2}\)

\(=\dfrac{36}{2}+\dfrac{3015}{2}\)

\(=\dfrac{3051}{2}\)

\(---\)

c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)

\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(---\)

d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)

\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)

\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)

\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)

\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)

\(=-\dfrac{25}{41}\)

#\(Toru\)

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BM

bé mèo miu

1 tháng 10 2023

cảm ơn nhiều nha vừa kịp giờ lun

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PT

Phi Thi Thu Hien

21 tháng 4 2022

Tính tổng sau đây bằng cách nhanh nhất

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\)

GIÚP MÌNH VỚI NHÉ MÌNH ĐAG CẦN GẤP.CHO MÌNH GỬI LỜI CẢM ƠN TRC NHÉ!

#Toán lớp 4

1

C

Chuu

21 tháng 4 2022

=\(\dfrac{64}{192}+\dfrac{32}{192}+\dfrac{16}{192}+\dfrac{8}{192}+\dfrac{4}{192}+\dfrac{2}{192}+\dfrac{1}{192}\)

= \(\dfrac{127}{192}\)

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