K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

23 tháng 4 2015

\(P=1+5+5^2+............+5^{2005}\)

\(5P=5+5^2+5^3+...........5^{2006}\)

\(5P-P=5^{2006}-1\)

\(P=\frac{5^{2006}-1}{4}\)

13 tháng 4 2017

dấu . là gì vậy bạn

13 tháng 4 2017

dấu nhân

8 tháng 5 2016

\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)

\(B=1-\frac{1}{2^{2016}}\)

phan kia tuong tu

21 tháng 7 2019

Bài 1:

1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3

\(\frac{11}{3}\)\(\frac{10}{3}\)- 3

\(\frac{11}{3}\)\(\frac{3}{10}\)- 3 

\(\frac{11}{10}\)- 3

\(\frac{-19}{10}\)

2) \(\frac{5}{6}\):  \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)

\(\frac{5}{6}\) . \(\frac{52}{3}\)\(\frac{5}{6}\). 47\(\frac{1}{3}\)

\(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))

\(\frac{5}{6}\).( -30)

= -25

21 tháng 7 2019

mách mình mấy câu kia với

22 tháng 3 2017

\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\) 

\(A=\frac{1105}{28}.\)\(\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{9+\frac{9}{7}-\frac{9}{11}+\frac{9}{1001}-\frac{9}{13}}\)

\(A=\frac{1105}{28}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}\)

\(A=\frac{1105}{28}.\frac{3}{9}\)

\(A=\frac{1105}{84}\)

b)\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)

Đặt \(A=1+2+2^2+2^3+...+2^{2012}\)

Suy ra \(2.A=2+2^2+2^3+2^4+...+2^{2013}\)

Khi đó \(2.A-A=2^{2013}-1\)hay \(A=2^{2013}-1\)

Do đó : \(M=\frac{A}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{1}{2}\)

          Vậy \(M=\frac{1}{2}\)

16 tháng 4 2018

gửi lắm thế m

13 tháng 12 2018

\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)

\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)

\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)

\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)

\(2A=2(1+2^2+2^3+...+2^{2012})\)

\(2A=2+2^2+2^3+...+2^{2013}\)

\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)

\(\Rightarrow A=2^{2013}-1\)

\(\text{Quay lại bài toán,ta có :}\)

\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)

14 tháng 4 2019

c)  \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\) 

\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\) 

\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=2\left(1-\frac{1}{16}\right)\) 

\(=2.\frac{15}{16}\) 

\(=\frac{15}{8}\) 

Vậy A=\(\frac{15}{8}\)

14 tháng 4 2019

a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}=\frac{297}{100}\)