Ta có: \(S=3+3^2+3^3+\cdots+3^{2024}\)

\(=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\cdots+\left(3^{2022}+3^{2023}+3^{2024}\right)\)

\(=12+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+\cdots+3^{2022}\left(1+3+3^2\right)\)

\(=12+13\left(3^3+3^6+\cdots+3^{2022}\right)\)

=>S không chia hết cho 13

b, = \(\frac{3^9\cdot3\cdot11+15\cdot3\cdot3^8}{16\cdot3^9}\)

= \(\frac{3^9\cdot33+15\cdot3^9}{16\cdot3^9}\)

= \(\frac{3^9\cdot\left(33+15\right)}{3^9\cdot16}\)

=\(\frac{48}{16}=3\)

a, =\(\frac{\left(3\cdot4\right)^5}{4^3\cdot3^4}\)= \(\frac{3^4\cdot3\cdot4^3\cdot4^2}{4^3\cdot3^4}\)=  3 * 4^2 = 3 * 16 = 48

b,    (3¹⁰*11+3⁸*45)/16*3⁹=3^10*11+3^10*5/16*3^9=3^10*16/3^9*16=3

a,   12^5/64*3^4=2^10*3^5/2^6*3^4=2^4*3=48

\(\Rightarrow3S=3^2+3^3+3^4+...+3^{2023}\)

trừ vế với vế ta được :

\(3S-S=3^{2023}-3\)

\(\Rightarrow2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

Giúp mình giải bài này nha

S=3+3^2+3^3+...+3^2022

3S=3.(3+3^2+3^3+...+3^2022)

3S=3^2+3^3+3^4+...+3^2023

⇒3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

⇒2S=3^2023-3

⇒S=3^2023-3 / 2

S=3+3^2+3^3+...+3^2022

=>3S=3^2+3^3+3^4+...+3^2023

=>3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

=>2S=3^2023-3

=>S=\(\dfrac{3^{2023}-3}{2}\)

Vậy S=​\(\dfrac{3^{2023}-3}{2}\)​

Ta có:

3²⁰¹⁸ = (3⁴)⁵⁰⁴ x 3² = (...1)⁵⁰⁴ x 9 = ...1 x9=....9

Vậy chữ số tận cùng là 9 => x=9

x=9

k nha

a)   \(x^3=216\)

=>  \(x^3=6^3\)

=>  \(x=6\)

ta có : 

\(A=3^1+3^2+3^3+3^4+...+3^{199}\)

\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)

\(2A=3^{200}-3^1\)

\(A=\frac{3^{200}-3}{2}\)

=))

Đặt \(A=3^1+3^2+3^3+...+3^{199}\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)

Lấy 3A trừ A theo vế ta có : 

\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)

\(2A=3^{200}-1\)

\(A=\frac{3^{200}-1}{2}\)

Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)