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![](https://rs.olm.vn/images/avt/0.png?1311)
x là số chắn
A=(-1)^n.3^n
A+3A=4A=1+(-1)^n.3^(n+1)
với x chẵn
A= [3^(x+1)+1]/4 vô nghiệm nguyên đề sai
![](https://rs.olm.vn/images/avt/0.png?1311)
tử số = 2^10 . 3^8 - 2^10 . 3^9 = 2^10 . 3^8 . (1-3)
mẫu số = 2^10 . 3^8 + 3^8 . 2^10 . 5 = 2^10 . 3^8 . (1+5)
Đ/S: \(\frac{1-3}{1+5}\)=\(\frac{-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 4+2^2+2^3+....+2^2015
\(\Rightarrow\)2A=8+2^3+2^4+...+2^2016
\(\Rightarrow\) 2A-A=8+2^3+2^4+....+2^2016 - 4 - 2^2 - 2^3 -.....- 2^2015
\(\Rightarrow\)A=8+2^2016 - 4 - 2^2
\(\Rightarrow\)A=2^2016
Vậy A là lũy thừa của 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề :
\(A=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2004}}\)
\(2A-A=\left(1+2+\frac{1}{2}+...+\frac{1}{2^{2004}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2005}}\right)\)
\(A=2-\frac{1}{2^{2005}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= \(6.\frac{4}{6}-3.\frac{4}{6}-2.\frac{-2}{3}+4\) \(=4-2+\frac{4}{3}+4=6\frac{4}{3}\)
\(6.\left(-\frac{2}{3}\right)^2-3.\left(-\frac{2}{3}\right)^2-2:\left(-\frac{3}{2}\right)+4\)\(\)\(\)
\(=6.\frac{4}{9}-3.\frac{4}{9}-2.\left(-\frac{2}{3}\right)+4\)
\(=\frac{4}{9}.\left(6-3\right)-2.\frac{-2}{3}+4\)
\(=\frac{4}{9}.3-2.\left(-\frac{2}{3}\right)+4\)
\(=\frac{2}{3}.2-2.\left(-\frac{2}{3}\right)+4\)
\(=-\frac{2}{3}.\left(-2-2\right)+4\)
\(=-\frac{2}{3}.\left(-4\right)+4\)
\(=\frac{8}{3}+\frac{12}{3}\)
\(=\frac{16}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=1+2+22+..............+22017
\(\Rightarrow\)2A =2+22+23+.............+22018
\(\Rightarrow\)2A -A = (2+22+23+............+22018) -(1+2+22 +...............+22017)
\(\Rightarrow\)A= 22018 -1
Lại có :A = ( 23 )672 .22 -1 =(7+1)672 .22 -1= ( B(7) +1).22 -1 =22 .B(7) +22-1=22 .B(7)+3
Vây A chia 7 dư 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Công thức tổng quát: }\frac{1}{1+2+3+...+n}=\frac{2}{\left(n+1\right).n}\)
bạn thay vào òi làm tiếp ,phần tiếp theo dễ thui
![](https://rs.olm.vn/images/avt/0.png?1311)
1: so sánh 2016/2017+2017/2018
vì 2016/2017 > 1/2017 >1/2018 =
> 2016/2017+2017/2018 >1/2018+2017/2018=1
vậy .....
Ta có : \(C=\frac{1}{2}+\left(-\frac{2}{3}\right)+\left(-\frac{2}{3}\right)^2+\left(-\frac{2}{3}\right)^3+......+\left(-\frac{2}{3}\right)^{2018}\)
\(\Rightarrow C=\frac{1}{2}-\left(\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+.....+\left(\frac{2}{3}\right)^{2018}\right)\)
Đặt \(\Rightarrow A=\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+.....+\left(\frac{2}{3}\right)^{2018}\)
\(\Rightarrow\frac{2}{3}A=\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^4+.....+\left(\frac{2}{3}\right)^{2019}\)
\(\Rightarrow A-\frac{2}{3}A=\frac{2}{3}-\frac{2}{3}^{2019}\)
\(\Rightarrow\frac{1}{3}A=\frac{2}{3}-\left(\frac{2}{3}\right)^{2019}\)
=> A = \(\left(\frac{2}{3}-\left(\frac{2}{3}\right)^{2019}\right).3\)
=> A = 2 - \(\frac{2^{2019}}{3^{2018}}\)