Question

[Yumani Jeng]

\(^{2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}=2^{2019}-8}.\)8 .Tìm số nguyên x .

Answer 1

[Trí Tiên]

Ta có : \(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8\)

\(\Leftrightarrow2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8\) (1)

Đặt : \(A=1+2+2^2+...+2^{2015}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2016}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)

\(\Rightarrow A=2^{2016}-1\)

Khi đó (1) trở thành :

\(2^x\left(2^{2016}-1\right)=2^{2019}-2^3\)

\(\Leftrightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)

\(\Leftrightarrow2^x=2^3\left(2^{2016}-1\ne0\right)\)

\(\Leftrightarrow x=3\)

Vậy : \(x=3\)

Answer 2

[Nguyễn Chí Tiến]

vậy x=3 nhé