2:

a: =>2(x+1)=26

=>x+1=13

=>x=12

b: =>(6x)^3=125

=>6x=5

=>x=5/6(loại)

c: =>\(7\cdot3^x\cdot\dfrac{1}{3}+11\cdot3^x\cdot3=318\)

=>3^x=9

=>x=2

d: -2x+13 chia hết cho x+1

=>-2x-2+15 chia hết cho x+1

=>15 chia hết cho x+1

=>x+1 thuộc {1;3;5;15}

=>x thuộc {0;2;4;14}

e: 4x+11 chia hết cho 3x+2

=>12x+33 chia hết cho 3x+2

=>12x+8+25 chia hết cho 3x+2

=>25 chia hết cho 3x+2

=>3x+2 thuộc {1;-1;5;-5;25;-25}

mà x là số tự nhiên

nên x=1

1: 

a: Đặt A=2^2024-2^2023-...-2^2-2-1

Đặt B=2^2023+2^2022+...+2^2+2+1

=>2B=2^2024+2^2023+...+2^3+2^2+2

=>B=2^2024-1

=>A=2^2024-2^2024+1=1

c: \(=\dfrac{3^{12}\cdot2^{11}+2^{10}\cdot3^{12}\cdot5}{2^2\cdot3\cdot3^{11}\cdot2^{11}}=\dfrac{2^{10}\cdot3^{12}\left(2+5\right)}{2^{13}\cdot3^{12}}\)

\(=\dfrac{7}{2^3}=\dfrac{7}{8}\)

\(2n^3-6=10\Rightarrow2n^3=16\Rightarrow n^3=8=2^3\Rightarrow n=2\\ 2n^2-8=10\Rightarrow2n^2=18\Rightarrow n^2=9\Rightarrow\left[{}\begin{matrix}n=3\\n=-3\end{matrix}\right.\)

a: \(\Leftrightarrow n^3=8\)

hay n=2

b: \(\Leftrightarrow n^2=9\)

hay \(n\in\left\{3;-3\right\}\)

2n + 10 ⋮ 2n - 3 (n \(\in\) Z)

2n - 3 + 13 ⋮ 2n - 3

13 ⋮ 2n - 3

2n - 3 \(\in\) Ư(13)  ={-13; -1; 1; 13}

n \(\in\) {-5; 1; 2; 8}

2n + 10 ⋮ 2n - 3 (n \(\in\) Z)

2n - 3 + 13 ⋮ 2n - 3

             13 ⋮ 2n - 3

2n - 3 \(\in\) Ư(13) = {-13; -1; 1; 13}

n \(\in\) {-5; 1; 2; 8}

20a = 20/1.21 + 20/2.22+ ... + 20/80.100

= 1-1/21 + 1/2 - 1/22 +...+ 1/80 - 1/100

= 1  + 1/2 + 1/3 +... + 1/19 + 1/20 - 1/81 - 1/82 -.... - 1/100

80b = 80/1.81 + 80/2.82 + 80/3.83 +... + 80/20.100

= 1 - 1/81+ 1/2 - 1/83 +...+ 1/20 - 1/100

=> 20a = 80b

=> a/b = 4 

\(S=2+2.2^2+3.2^3+...+2016.2^{2016}\)

\(2S=2^2+2.2^3+3.2^4+...+2016.2^{2017}\)

\(2S-S=S=\text{​​}\text{​​}\text{​​}\text{​​}2^2+2.2^3+3.2^4+...+2016.2^{2017}-2-2.2^2-3.2^3-...-2016.2^{2016}\)

\(S=2\left(0-1\right)+2^2\left(1-2\right)+2^3\left(2-3\right)+...+2^{2016}\left(2015-2016\right)+2^{2017}.2016\)

\(S=-\left(2+2^2+2^3+...+2^{2016}\right)+2^{2017}.2016\)

\(\)Đặt \(A=2+2^2+2^3+...+2^{2016}\)

\(2A=2^2+2^3+2^4+...+2^{2017}\)

\(2A-A=A=2^2+2^3+2^4+...+2^{2017}-2-2^2-2^3-...-2^{2016}\)

\(A=2^{2017}-2\)

Thay vào S ta được:
\(S=-2^{2017}+2+2^{2017}.2016\)

\(S=2^{2017}.2015+2\)

Ta có \(S+2013=2^{2017}.2015+2+2013\)

\(S+2013=2^{2017}.2015+2015\)

\(S+2013=2015\left(2^{2017}+1\right)\)

Suy ra \(S+2013⋮2^{2017}+1\)

Vậy \(S+2013⋮2^{2017}+1\) (đpcm)

cái này dễ lắm lun

\(10n+10=10n+5+5=5\left(2n+1\right)+5⋮2n+1\)

\(=>2n+1\inƯ\left(5\right)\\ Ư\left(5\right)=\left\{1;-1;5;-5\right\}\\ =>n=\left\{0-1;4;-3\right\}\)