BCNN(1;2;3;4;5;6;7;8;9;10)=2520 nha bn

a, 

A = 2 + 2² + 2³ +...+2¹⁰

A = (2 + 2² ) + (2³ +2⁴ ) + ...+ (2⁹ + 2¹⁰ )

A = 2 ( 1+2 ) + 2³(1+2 ) + ...+ 2⁹(1+2)

A = 2 .3 + 2³ .3 + ...+2⁹.3

A = 3 ( 2+ 2³ + ...+ 2⁹ ) \(⋮\) 3 3

Vậy A \(⋮\) 3

b, A = 2 + 2² + 2³ +...+2¹⁰

A = ( 2 + 2² + 2³ + 2⁴ + 2⁵ ) + ( 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰ )

A =  2 ( 1+2+2² + 2³ + 2⁴ ) + 2⁶(1+2+2² + 2³ + 2⁴)

A = 2 . 31 + 2⁶ .31

A = 31(2+2⁶ ) \(⋮\) 31

vậy A \(⋮\) 31

d , A = 2 + 2² + 2³ +...+2¹⁰  

lam nhanh minh  k cho nha

\(a,4^x=64\)

\(4^x=4^3\)

\(x=3\)

\(b,3\cdot\left(5^x-16\right)=3^3\)

\(5^x-16=3^3:3=3^2\)

\(5^x-16=9\)

\(5^x=9+16=25\)

\(5^x=5^2\)

\(x=2\)

\(=\frac{9^3.8^3.9^2.6^2}{9^4.3^4.4^4}=\frac{9^5.4^3.2^3.2^2.3^2}{9^4.4^4.3^4}\)\(=\frac{9.2^3.2^2}{4.3^2}=2^3=8\)

\(=\frac{9^3.8^3.9^2.6^2}{9^4.12^4}=\frac{9.4^3.2^3.3^2.2^2}{3^4.4^4}=\frac{9.2^3.2^2}{3^2.4}\) =2³=8  

Đề sai thì phải ! Học Lớp 7 mới giải xong bài này !

\(\frac{1}{9}\cdot27^n=3^n\)

\(\frac{1}{9}\cdot\left(3^3\right)^n=3^n\)

\(\frac{1}{9}\cdot3^{3n}=3^n\)

\(\frac{1}{9}=3^n\text{ : }3^{3n}\)

\(\frac{1}{9}=3^{-2n}\)

\(\frac{1}{3^2}=\frac{1}{3^{2n}}\)

\(\Rightarrow\text{ }3^{2n}=3^2\)

\(3^{2n}-3^2=0\)

\(3\left(3^{2n-1}-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3=0\text{ ( Vô lí ) }\\3^{2n-1}-3=0\end{cases}}\)       \(\Rightarrow\text{ }3^{2n-1}=3\)          \(\Rightarrow\text{ }2n-1=1\) \(\Rightarrow\text{ }2n=2\) \(\Rightarrow\text{ }n=1\)

                Vậy \(n=1\)

\(\frac{1}{9}\cdot27^n=3^n\)

\(\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)

\(\frac{3^{3n}}{3^2}=3^n\)

\(3^{3n}=3^2\cdot3^n\)

\(3^{3n}=3^{n+2}\)

\(\Rightarrow\text{ }3n=n+2\)

\(3n-n=2\)

\(2n=2\)

\(n=2\text{ : }2\)

\(n=1\)