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![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2+22+23+...+22017 (1)
2A=2+22+23+24+...+22018 (2)
Lấy (2) - (1) ta có:
2A - A=(2+22+23+24+...+22018)-(1+2+22+23+...+22017)
A=2+22+23+24+...+22018-1-2-22-23-...-22017
A=22018-1
Mà B=22018-1 =>A=B
b) ta có: B=20172
B=(2016+1).2017=2016.2017+2017
A=2016.2018
A=2016.(2017+1)=2016.2017+2016
Vì 2016<2017=>A<B
mình nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,5mũ 36=(5mũ3)mũ12=125 mũ12
11^24=(11^2)12=121^12
vì 121<125 nên 5^36>11^24
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{2^{2017}+2}{2^{2017}+3}=1-\frac{1}{2^{2017}+3}\)
\(B=\frac{2^{2017}+1}{2^{2017}+2}=1-\frac{1}{2^{2017}+2}\)
Vì \(\frac{1}{2^{2017}+3}< \frac{1}{2^{2017}+2}\) nên \(1-\frac{1}{2^{2017}+3}>1-\frac{1}{2^{2017}+2}\)
hay A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(2A=2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)
\(\Rightarrow A=2^{2018}-1< 2^{2018}=B\)
Vậy A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ko ghi đề
\(2A=2+2^2+...+2^{101}\\ 2A-A=2^{101}-1\\ =>A=2^{101}-1\)
Mấy cái khác cg lm như v (b thì 3b)
Nhớ đúng mk nhá
Ta có : \(A=1+2+2^2+...+2^{2017}\)(1)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2018}\)(2)
Lấy (2) trừ (1) ta có :
\(\Rightarrow A=2^{2018}-1\)
\(\Rightarrow A< B\). Vì \(B=2^{2018}\)
A = 1+2+22+23+.....+22017
2A = 2(1+2+22+23+.....+22017) = 2+22+23+24+.....+22018
2A - A = 2+22+23+24+.....+22018- (1+2+22+23+.....+22017)
=> A = 2+22+23+24+.....+22018-1-2-22-23-.....-22017
A =22018-1 < 22018
Vậy A < B