\(A=1+2+2^2+2^3+2^4+...+2^{49}\)

\(2A=2+2^2+2^3+2^4+..+2^{50}\)

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

\(\Rightarrow A=2^{50}-1\)

\(9.9^{13}=9^{1+13}=9^{14}\)

\(8^{13}\div8=8^{13-1}=8^{12}\)

\(2^{13}.2=2^{13+1}=2^{14}\)

\(8^2\div8=8^{2-1}=8^1\)

\(9^{1+13}=9^{14}\)

\(8^{13-1}=8^{12}\)

\(2^{13+1}=2^{14}\)

\(8^{2-1}=8\)

tích mik nha

mik tích lại

a) 2012 x 8 - 8 x 2011

= 8 x (2012 - 2011)

= 8 x 1 

= 8

b) \(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}\)

\(=\frac{7}{8}.\frac{8}{3}-\frac{5}{8}.\frac{8}{3}\)

\(=\frac{8}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)\)

\(=\frac{8}{3}.\frac{2}{8}\)

\(=\frac{16}{6}=\frac{8}{3}\)

1,

a,\(2012.8-8.2011\)

\(=8\left(2012-2011\right)\)

\(=8.1\)

\(=8\)

b.\(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}\)

\(=2\frac{2}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)\)

\(=\frac{8}{3}.\frac{1}{4}\)

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

thanks bro ! :)

tích cho mk đi bro

 8 ^ 7

6 ^5

\(8^2\)*\(8^5\)=\(8^{2+5}\)=\(8^7\)

\(6^7\):\(6^2\)=\(6^{7-2}\)=\(6^5\)

Kết quả là A>B bạn nhé

a.Ta có: n² +n + 1

=n.(n+1) +1

Vì n+1 chia hết cho n+1 => n.(n+1) chia hết cho n+1

Để n.(n+1)+1 chia hết cho n+1 => 1 chia hết cho n+1.

=> n+1 thuộc Ư(1)

Mà n thuộc N => n=1

Vậy n=1.

a) Ta có : \(n^2+n+1=n\left(n+1\right)+1\)

\(\Rightarrow n^2+n+1⋮n+1\Leftrightarrow1⋮n+1\) ( vì \(n\left(n+1\right)⋮n+1\))

                       \(\Leftrightarrow n+1\inƯ\left(1\right)=\left\{1\right\}\) ( vì \(n\inℕ\))

                   \(\Rightarrow n=1-1=0\)

Vậy  \(n=0\)

Làm tương tự với các câu còn lại.

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)

A >1,B<1 suy ra A>B