Question

[Nguyễn Hà Linh]

2 mũ x+2 mũ x+2 mũ x+1+....+2 mũ x+2015=2 mũ 2019-8

cứu tui với

gấp lắm ạ

Answer 1

[Lưu Nguyễn Hà An]

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

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

Đặt : $A=1+2+2^2+...+2^{2015}$

$\Rightarrow 2A=2+2^2+2^3+...+2^{2016}$

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

$\Rightarrow A=2^{2016}-1$

Khi đó (1) trở thành :

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

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

$\iff 2^x=2^3\left(2^{2016}-1\ne 0\right)$

$\iff x=3$

Vậy : $x=3$

Answer 2

[Lưu Nguyễn Hà An]

$2^x+2^{x+1}+...+2^{x+2015}=2^{2019}-8$

$\to 2^x.1+2^x.2+....+2^x.2^{2015}=2^{2019}-8$

$\to 2^x.(1+2+...+2^{2015})=2^{2019}-8$

Đặt:

$A=1+2+...+2^{2015}$

$2A=2.(1+2+...+2^{2015})$

$2A=2+2^2+...+2^{2016}$

$2A-A=(2+2^2+...+2^{2016})-(1+2+...+2^{2015})$

$A=2+2^2+...+2^{2016}-1-2-...-2^{2015}$

$A=2^{2016}-1$

Nên:

$2^x.(1+2+...+2^{2015})=2^{2019}-8$

$\to 2^x.(2^{2016}-1)=2^{2019}-8$

$\to 2^x=(2^{2019}-8):(2^{2016}-1)$

$\to 2^x=\frac{2^{2019}-8}{2^{2016}-1}$

$\to 2^x=\frac{2^3.(2^{2016}-1)}{2^{2016}-1}$

$\to 2^x=2^3$

$\to x=3$

Vậy $x=3.$