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24 tháng 5 2017

\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+....+\left(-\dfrac{1}{7}\right)^{2017}\\ =1+-\dfrac{1}{7}+\dfrac{1}{7^2}+-\dfrac{1}{7^3}+.....+-\dfrac{1}{7^{2017}}\\ =\left(1+\dfrac{1}{7^2}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{2016}}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{2017}}\right)\)

rồi bạn tính 2 về rồi trừ ra là xng nhé

24 tháng 5 2017

bài này hình như ra 7/8

12 tháng 12 2021

\(B=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2018}\)

\(\Rightarrow-\dfrac{1}{7}B=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}\)

\(\Rightarrow-\dfrac{1}{7}B-1=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}-\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^1-\left(-\dfrac{1}{7}\right)^2-...-\left(-\dfrac{1}{7}\right)^{2018}\)

\(\Rightarrow-\dfrac{8}{7}B=\left(-\dfrac{1}{7}\right)^{2019}-1\)

\(\Rightarrow B=\left[\left(-\dfrac{1}{7}\right)^{2019}-1\right]:\left(-\dfrac{8}{7}\right)\)

12 tháng 12 2021

\(B=1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow7B=7-1+\dfrac{1}{7}-\dfrac{1}{7^2}+...-\dfrac{1}{7^{2016}}+\dfrac{1}{7^{2017}}\\ \Rightarrow7B+B=6+\dfrac{1}{7}-\dfrac{1}{7^2}+...+\dfrac{1}{7^{2017}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow8B=7+\dfrac{1}{7^{2018}}=\dfrac{7^{2019}+1}{7^{2018}}\\ \Rightarrow B=\dfrac{7^{2019}+1}{8\cdot7^{2018}}\)

25 tháng 11 2021

:) 

undefined

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...
Đọc tiếp

tính

a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)

b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)

c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)

e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)

h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)

3
7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

7 tháng 10 2017

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

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

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

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

12 tháng 8 2017

\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\\ S=\dfrac{\left(-1\right)^0}{7^0}+\dfrac{\left(-1\right)^1}{7^1}+\dfrac{\left(-1\right)^2}{7^2}+...+\dfrac{\left(-1\right)^{2017}}{7^{2017}}\\ S=\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\\ -7S=\dfrac{-7}{7^0}+\dfrac{7}{7^1}+\dfrac{-7}{7^2}+...+\dfrac{7}{7^{2017}}\\ -7S=\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\\ -7S-S=\left[\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\right]+\left(\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\right)\\ -8S=\left(-7\right)+\dfrac{-1}{2017}\\ -8S=-\left(7+\dfrac{1}{2017}\right)\\ 8S=7+\dfrac{1}{2017}\\ S=\dfrac{7+\dfrac{1}{2017}}{8}\)

Vậy ...

10 tháng 4 2018

https://hoc24.vn/hoi-dap/question/266859.html

25 tháng 12 2017

S= \(\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\)

\(\left(-\dfrac{1}{7}\right)S=\left(-\dfrac{1}{7}\right)\left(-\dfrac{1}{7}+-\dfrac{1^2}{7}+..+-\dfrac{1^{2007}}{7}\right)\)

= \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\)

=>\(-\dfrac{1}{7}S-S=\) \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\) \(-\)\(\left(1+-\dfrac{1}{7}+-\dfrac{1^2}{7}+...+-\dfrac{1^{2007}}{7}\right)\)

=> \(-\dfrac{1}{7}S=\) \(\dfrac{-1^{2008}}{7}-1\)

=> S= \(\dfrac{-1^{2008}}{7}-1\) : \(\dfrac{-1}{7}\)

9 tháng 5 2017

mk ko chép đề đâu nha

\(S=1+\dfrac{-1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)

đặt \(7S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\)

=>\(7S+S=\left(7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\right)+\left(1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\right)\)

=>\(8S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)

=>\(8S=7+\left(-1+1\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+...+\left(\dfrac{1}{7^{2015}}-\dfrac{1}{7^{2015}}\right)+\dfrac{1}{7^{2016}}\)

=> \(8S=7+\dfrac{1}{7^{2016}}\)

\(\Rightarrow S=\dfrac{7+\dfrac{1}{7^{2016}}}{8}\)

9 tháng 5 2017

Gỉa sử : \(-\dfrac{1}{7}=a\)

Thay vào S ,có :

\(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) (1)

=> a.S = a( \(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) )

= \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) (2)

Lấy (2) - (1) ,CÓ :

aS-S=( \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) ) - ( \(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) ) aS-S= \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) - \(1-a-a^2-.........-a^{2016}\)

aS-S = a2017 -1 => S(a-1) = a2017 -1

=> S = \(\dfrac{a^{2017}-1}{a-1}\)

Thay a= -1/7 vào S = \(\dfrac{a^{2017}-1}{a-1}\) ,có :

S = \(\dfrac{\left(\dfrac{-1}{7}\right)^{2017}-1}{-\dfrac{1}{7}-1}=\dfrac{\left(-\dfrac{1}{7}\right)^{2017}}{-\dfrac{8}{7}}\)

\(=\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot\dfrac{4}{7}\cdot\dfrac{3}{7}\cdot\dfrac{2}{7}\cdot\dfrac{1}{7}\cdot\dfrac{0}{7}\cdot...\cdot\dfrac{-2007}{7}\)

=0

30 tháng 11 2021

2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)