3 mũ 11:52 là số nào vậy bạn

\(Q=1+3+3^2+3^3+3^4+...+3^{11}\)

\(3Q=3+3^2+3^3+3^4+3^5+...+3^{12}\)

\(3Q-Q=\left(3+3^2+3^3+3^4+3^5+...+3^{12}\right)-\left(1+3+3^2+3^3+3^4+...+3^{11}\right)\)

\(2Q=3^{12}-1\)

\(Q=\frac{3^{12}-1}{2}\)

Sai rroif mk tưởng tính Q

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

a)3¹x3²x3³x........x3¹⁰⁰

=3^{1+2+3+4+...+100}

=3^{(100+1)x(100-1+1):2}

=3^101x100:2

=3⁵⁰⁵⁰

Bài b mình không biết làm

thank nha

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

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

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

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

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

=))

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

xích ma 3x chạy từ 1 tới 199 kết quả là \(^{\text{1,328069944 nhân}10^{95}}\)

G=1-3+3²-3³+3⁴-...-3⁹⁹+3¹⁰⁰

3G=3-3²+3³-3⁴+3⁵-....-3¹⁰⁰+3¹⁰¹

3G+G=(3-3²+3³-3⁴+3⁵-....-3¹⁰⁰+3¹⁰¹)+(1-3+3²-3³+3⁴-...-3⁹⁹+3¹⁰⁰)

4G = 3¹⁰¹+1

G=\(\frac{3^{101}+1}{4}\)

\(A=\)\(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

\(3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\)

\(4A=-1-\frac{1}{3^{51}}\)

\(A=\frac{-1-\frac{1}{3^{51}}}{4}\)

k cho mik nha