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KN
28 tháng 3 2019
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^{^2}}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow2A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow A=\left(1-\frac{1}{3^{99}}\right)\div2\)
25 tháng 2 2018
Ta có :
\(S=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)
\(\Leftrightarrow\)\(3S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Leftrightarrow\)\(3S-S=\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\right)\)
\(\Leftrightarrow\)\(2S=\frac{1}{3}-\frac{1}{3^9}\)
\(\Leftrightarrow\)\(2S=\frac{3^8-1}{3^9}\)
\(\Leftrightarrow\)\(S=\frac{3^8-1}{2.3^9}\)
Ở đây mk chỉ ghi \(...\) cho nhanh nếu bạn làm vào vở thì ghi đầy đủ ra nhé
2 tháng 4 2017
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...................+\dfrac{1}{3^{2016}}\)
\(3A=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...............+\dfrac{1}{3^{2016}}\right)\)
\(3A=\dfrac{3}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+.................+\dfrac{3}{3^{2016}}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+......................+\dfrac{1}{3^{2015}}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{2015}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+.............+\dfrac{1}{3^{2016}}\right)\)
\(A=1-\dfrac{1}{3^{2016}}\)
\(A=\dfrac{3^{2016}-1}{3^{2016}}\)
~~ Chúc bn học tốt ~~
2 tháng 4 2017
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\)
\(\dfrac{1}{3}A=\dfrac{1}{3}.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\right)\)
\(\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2017}}\)
\(\dfrac{1}{3}A-A=\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2017}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\right)\)
\(-\dfrac{2}{3}A=\dfrac{1}{3^{2017}}-\dfrac{1}{3}\)
\(A=\dfrac{\dfrac{1}{3^{2017}}-\dfrac{1}{3}}{-\dfrac{2}{3}}\).
15 tháng 7 2017
a/ \(8^5=\left(2^3\right)^5=2^{15}\)và \(32^3=\left(2^5\right)^3=2^{15}\Rightarrow8^5=32^3\)
b/ \(27^4=\left(3^3\right)^4=3^{12}\) và \(9^6=\left(3^2\right)^6=3^{12}\Rightarrow27^4=9^6\)
c/ \(23^{17}-23^{16}=23^{16}\left(23-1\right)=22.23^{16}\)
\(23^{16}-23^{15}=23^{15}\left(23-1\right)=22.23^{15}\)
\(\Rightarrow22.23^{16}>22.23^{15}\Rightarrow23^{17}-23^{16}>23^{16}-23^{15}\)
d/ \(\frac{3^{2015}+1}{3^{2016}}=\frac{1}{3}+\frac{1}{3^{2016}}\) và \(\frac{3^{2016}+1}{3^{2017}+1}=\frac{3^{2017}+3}{3\left(3^{2017}+1\right)}=\frac{3^{2017}+1+2}{3\left(3^{2017}+1\right)}=\frac{1}{3}+\frac{2}{3}.\frac{1}{3^{2017}+1}\)
\(\frac{1}{3^{2016}}>\frac{1}{3^{2017}}>\frac{1}{3^{2017}+1}>\frac{2}{3}.\frac{1}{3^{2017}+1}\)
\(\Rightarrow\frac{3^{2015}+1}{3^{2016}}>\frac{3^{2016}+1}{3^{2017}+1}\)
Câu cuối phân tích tương tự
=-10/3