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

10 tháng 4 2018

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

25 tháng 11 2021

:) 

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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 ...

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

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

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

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

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

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

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

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

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

29 tháng 10 2021

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

30 tháng 11 2021

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