\(3+3^2+...+3^{2022}\)

\(=\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)

\(=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{2020}\cdot\left(1+3+9\right)\)

\(=3\cdot13+3^4\cdot13+...+3^{2020}\cdot13\)

\(=13\cdot\left(3+3^4+...+3^{2020}\right)\) ⋮ 13 

Vậy.... 

2n+3 chia hết cho n-2

=> 2(n-2)+7 chia hết cho n-2

=> 7 chia hết cho n-2

Hay n-2 thuộc Ư(7)={1;7;-1;-7}

=> n thuộc { 3;9;1;-5}

Ta có: 2n+3 chia hết n-2; n-2 chia hết cho n-2

=> (2n+3) -2 x (n-2) chia hết cho n-2

=> 7 chia hết cho n-2

=> n-2 thuộc Ư(7)={ -1 ; -7 ; 1 ; 7 }

=> n thuộc { 1 ; -5 ; 3 ; 9 }

Vậy ....

Lời giải:

$(\frac{1}{24}-\frac{5}{16}): \frac{-3}{8}+1$

$=\frac{-13}{48}.\frac{-8}{3}+1=\frac{13}{18}+1=\frac{31}{18}$

=(4/96-30/96) : (-3/8) + 1 

=(-3/8) : (-3/8) + 1

=1+1

=2

\(x\in B\left(5\right)=\left\{0;5;10;15;20;25;30;35;40;...\right\}\)

Mà \(20\le x\le36\) nên \(x=20;25;30;35\)

Ta có:

\(2^{200}.2^{100}=\left(2^2\right)^{100}.2^{100}=4^{100}.2^{100}=\left(4.2\right)^{100}=8^{100}\)

\(3^{100}.3^{100}=\left(3.3\right)^{100}=9^{100}\)

Vì \(8< 9\) nên \(8^{100}< 9^{100}\)

Vậy \(2^{200}.2^{100}< 3^{100}.3^{100}\)

\(#WendyDang\)

\(2^{200}\cdot2^{100}=2^{300}=(2^3)^{100}=8^{100}\\3^{100}\cdot3^{100}=(3\cdot3)^{100}=9^{100}\)

Vì \(8< 9\) nên \(8^{100}< 9^{100}\)

hay \(2^{200}\cdot2^{100}< 3^{100}\cdot3^{100}\)

\(1^{22}\times3^{22222}\)

\(1\times3^{22222}\)

\(3^{22222}\)

Hh

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18