dễ mà ngu thế mấy bọn não cá vàng

2^x(1+2+2²+2³+.........+2²⁰¹⁵)=2²⁰¹⁹-8

dặt 1+2+..........+2²⁰¹⁵ là a

2a=2x(1+2+...........+2²⁰¹⁵)

2a=2+2²+........+2²⁰¹⁶

2a-a=2²⁰¹⁶-1

a=2²⁰¹⁶-1

2^x(2²⁰¹⁶-1)=2³(2²⁰¹⁶-1)

2^x=2³

x=3

Ok bn là nhất

Ta có : 2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-82x+2x+1+2x+2+...+2x+2015=22019−8

\Leftrightarrow2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8⇔2x(1+2+22+...+22015)=22019−8 (1)

Đặt : A=1+2+2^2+...+2^{2015}A=1+2+22+...+22015

\Rightarrow2A=2+2^2+2^3+...+2^{2016}⇒2A=2+22+23+...+22016

\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)⇒2A−A=(2+22+23+...+22016)−(1+2+22+...+22015)

\Rightarrow A=2^{2016}-1⇒A=22016−1

Khi đó (1) trở thành :

2^x\left(2^{2016}-1\right)=2^{2019}-2^32x(22016−1)=22019−23

\Leftrightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)⇔2x(22016−1)=23(22016−1)

\Leftrightarrow2^x=2^3\left(2^{2016}-1\ne0\right)⇔2x=23(22016−1=0)

$\iff x=3$

Vậy : $x=3$

x = 3 nhé

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 : \(x=3\)

vậy x=3 nhé

2^x.[1+2+2^2+....+2^2015]­=2^2019-8

ta co A=1+2+2^2+....+2^2015

2A=2+2^2+2^3+....+2^2016

2A-A=2^2016-1

thay A vao ta co 

2^ x.[2^2016-1]=2^2016.2^3-2^3

2^x=2^3.[2^2016-1]:[2^2016-1]

2^x=2^3

=>x=3