Tính \(B=-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
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26 tháng 10 2020
A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
NN
26 tháng 10 2020
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
3 tháng 7 2017
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
5 tháng 12 2018
B=-4/5+4/52-4/53+...+4/5200
5B=-4+4/5-4/52+...+4/5201
5B+B=-4+4/5200
6B=-4x5200/5200+4/5200
6B=-4+4x5200/5200
Còn lại bạn tính nốt nha
NH
1
23 tháng 1 2017
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\)
\(2B=2\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\right)\)
\(2B=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{200}{2^{199}}\)
\(2B-B=\left(2+\frac{3}{2^2}+...+\frac{200}{2^{199}}\right)-\left(1+\frac{3}{2^3}+...+\frac{200}{2^{200}}\right)\)
.... đặt A=... giiả tiếp
HQ
Hà Quang Minh
Giáo viên
8 tháng 10 2023
a)
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)
c)
\(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)
DP
1 tháng 7 2017
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
\(-5B=4-\frac{4}{5}+\frac{4}{5^2}-....-\frac{4}{5^{199}}\)
\(-5B-B=4-\frac{4}{5^{200}}\)
\(-6B=\frac{4\left(5^{200}-1\right)}{5^{200}}\)
\(B=\frac{2\left(1-5^{200}\right)}{5^{200}.3}\)
Ta có:\(B=...\)
\(\Leftrightarrow5B=-4+\frac{4}{5}-\frac{4}{5^2}+...+\frac{4}{5^{199}}\)
\(\Leftrightarrow5B+B=-4+\frac{4}{5^{200}}\)
\(\Leftrightarrow6B=-4+\frac{4}{5^{200}}\Rightarrow B=\frac{-4+\frac{4}{5^{200}}}{6}=\frac{-2+\frac{2}{5^{200}}}{3}\)