(2x+1)(y-3)=12

Vì x;y là số tự nhiên => 2x+1;y-3 là số tự nhiên

                                 => 2x+1;y-3 E Ư(12)

Ta có bảng:

Vậy cặp số tự nhiên (x;y) cần tìm là: (0;15) ; (1;7)

(2x + 1)(y - 3) = 12

=> 2x + 1;y - 3 thuộc Ư(12)

vì x là stn => 2x + 1 là stn, ta có bảng

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

=> (1+2X-1)x (2x-1+1)/4=225

=> 2x+2x/4=225

=> 4x^2/4=225

=> x^2= 225

=> x=15

cái ^ là mũ nha bạn

chúc bn hok tốt

`Answer:`

a. Tổng: \([\left(2x-1\right)-1]:2+1=x\) số hạng

Ta có: \(1+3+5+7+9+...+\left(2x-1\right)=225\)

\(\Rightarrow x.\left(2x-1+1\right):2=225\)

\(\Leftrightarrow2x^2:2=225\)

\(\Leftrightarrow x^2=225\)

\(\Leftrightarrow x=15\)

b. Mình sửa đề nhé: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}=2^{2019}-8\)

\(\Rightarrow2^x.\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8\)

Ta đặt \(K=1+2+2^2+...+2^{2015}\)

\(\Rightarrow2^x.K=2^{2019}-8\)

\(\Rightarrow2K=2.\left(1+2+2^2+...+2^{2015}\right)\)

\(\Rightarrow2K=2+2^2+2^3+...+2^{2015}+2^{2016}\)

\(\Rightarrow2K-K=\left(2+2^2+2^3+...+2^{2015}+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)

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

\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^{2019}-8\)

\(\Rightarrow2^{x+2016}-2^x=2^{2019}-2^3\)

\(\Rightarrow\hept{\begin{cases}x+2016=2019\\x=3\end{cases}}\Rightarrow x=3\)

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

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

2^x+2^x+1+2^x+2+2^x+3+....................+2^x+2015=2^2019-8

2^{x.(1+2+3+...............+2015)}=2²⁰¹¹

=>x.(1+2+3+........+2015)=2011

               Có 2015 số hạng

=>x.[(1+2015).2015:2]=2011

=>x.2031120=2011

=>x                     =\(\frac{2011}{2031120}\)

Vậy x=\(\frac{2011}{2031120}\)

Chúc bn học tốt