Question

so sanh A voi 1 biet A= 2^2019-(2^2018+2^2017+...+2^1+2^0)

Answer 1

\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)

Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)

\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)

\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)

\(\Rightarrow B=2^{2019}-1\)

\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)

\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)

Answer 2

đoạn cuối cùng bạn làm sai rồi

Answer 3

Mình nhầm ạ ~

\(2^{2019}-\left(2^{2019}-1\right)=2^{2019}-2^{2019}+1=1.\)