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Câu 2: A = \(^{1+2+2^2+2^{ }^3+...+2^{2017}}\)
2A = \(2+2^2+2^3+...+2^{2018}\)
Suy ra 2A - A =\(2^{2018}-1\) Do đó A < B
1. Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\Rightarrow a=2016t,b=2017t,c=2018t\)
\(\left(a-c\right)^3=\left(2016t-2018t\right)^3=\left(-2t\right)^3=-8t^3\)
\(8\left(a-b\right)^2\left(b-c\right)=8\left(2016t-2017t\right)^2\left(2017t-2018t\right)=8.\left(-t\right)^2.\left(-t\right)=-8t^3\)
Vậy \(\left(a-c\right)^3=8\left(a-b\right)^2\left(b-c\right)\)
Đặt \(A=\frac{2^{2017}+1}{2^{2018}+1}\Rightarrow2A=\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
\(B=\frac{2^{2018}+1}{2^{2019}+1}\Rightarrow2B=\frac{2^{2019}+2}{2^{2019}+1}=\frac{2^{2019}+1+1}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Vì \(2^{2019}+1>2^{2018}+1\Rightarrow\frac{1}{2^{2019}+1}< \frac{1}{2^{2018}+1}\)
\(\Rightarrow2A>2B\Rightarrow A>B\)
Đặt G=2^2017+2^2016+...+2+1
=>2G=2^2018+2^2017+...+2^2+2
=>G=2^2018-1
=>H=2^2018-2^2018+1=1
=>2018^H=2018
A > B
Vì A là 20182-20162 là bình phương lên sẽ lớn hơn là 1 x 2017
So sánh A và B, biết:
A = 20182 - 20162 và B = 2. 2017
Ta có:
A= 20182 - 20162
=2018. (2017 + 1) - 2016. (2017 - 1)
=2018. 2017 + 2018 - 2016. 2017 + 2016
=2017. (2018 - 2016) + 4034
=2017. 2 + 4034
_Vì 2017. 2 + 4034 > 2. 2017
=>20182 - 20162 > 2. 2017
=>A >B
\(M=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)
Gọi \(A=2018^{2019}+2018^{2018}+...+2018^2+2018\)
\(\Rightarrow2018A=2018^{2020}+2018^{2019}+...+2018^3+2018^2\)
\(\Rightarrow2018A-A=2018^{2020}-2018\)
\(\Rightarrow2017A=2018^{2020}-2018\)
\(\Rightarrow A=\left(2018^{2020}-2018\right)\div2017\)
\(\Rightarrow M=\left(2018^{2020}-2018\right)\div2017.2017+1\)
\(\Rightarrow M=2018^{2020}-2018+1\)
\(\Rightarrow M=2018^{2020}-2017\)
Ta có \(A=1+2+2^2+2^3+...+2^{2017}\)
Suy ra\(2.A=2+2^2+2^3+2^4+....+2^{2018}\)
Khi đó \(2A-A=2+2^2+2^3+2^4+....+2^{2018}-\left(1+2+2^2+2^3+....+2^{2017}\right)\)
Hay \(A=2^{2018}-1\)
Ta thấy \(A=2^{2018}-1\); \(B=2^{2018}-1\)nên \(A=B\)
Vậy \(A=B\)
A=22019-(22018+22017+...+21+20)
Đặt M =22018+22017+...+21+20
M=22018+22017+...+2+1
2M=22019+22018+...+22+2
2M-M=(22019+22018+...+22+2)-(22018+22017+...+2+1)
M=22019-1
Suy ra:A=22019-(22019-1)
A=22019-22019+1
A=1
Vậy A=1
Ta có : \(A=2^{2019}-\left(2^{2018}+2^{2017}+...+2^1+2^0\right)\)
Đặt \(B=2^0+2^1+...+2^{2017}+2^{2018}\\ \Rightarrow2B=2+2^2+...+2^{2019}\\ \Rightarrow2B-B=\left(2+2^2+...+2^{2019}\right)-\left(2^0+2^1+...+2^{2017}+2^{2018}\right)\\ \Rightarrow B=2^{2019}-2^0\\ \Rightarrow A=2^{2019}-\left(2^{2019}-2^0\right)\\ \Rightarrow A=2^0=1\)
Vậy A=1
A = B thì phải
no.A=B.ok