\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)

\(=\frac{2^2.3}{4}+3\)

\(=3+3=6\)

b)

\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)

\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)

a) \(A=4+2^2+2^3+2^4+....+2^{20}\)

\(\Rightarrow2A=2^3+2^3+2^4+.....+2^{21}\)

\(\Rightarrow2A-A=\left(2^3+2^3+2^4+....+2^{21}\right)-\left(2^2+2^3+2^4+...+2^{20}\right)\)

\(\Rightarrow A=2^3+2^{21}-\left(2^2+2^2\right)\)

\(\Rightarrow A=2^{21}\)

\(\text{Vì }2^{21}⋮2^7\Rightarrow A⋮128\)

b) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)

\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)

\(=3+3=6\)

cách làm :

câu B 

ta thấy  có 2 lần 2¹⁰ xuất hiện trên 1 phần tử

ta gộp lại như sau :

2¹⁰ x ( 13 + 65 ) cho dễ 

còn câu A 

ta không thể tóm gọn nên phải tính như bình thường 

A  = 649539 + 45  /  648

A = 1002 . 44444444444444444

hay 649584 / 648

B = 0 + 66560 / 2912

B  = 22 . 8571428571

hay 66560 / 2912

Đ/s : .........

nhé !

cho mình sửa lại câu B  

B  =  1024 x ( 13 + 65 ) / 2912

B =   79872 / 2912

\(A=\frac{72^3\times54^2}{108^4}=\frac{\left(2^3\times3^2\right)^3\times\left(2\times3^3\right)^2}{\left(2^2\times3^3\right)^4}=\frac{2^9\times3^6\times2^2\times3^6}{2^8\times3^{12}}=2^3=8\)

\(B=\frac{3^{10}\times11+3^{10}\times5}{3^9\times2^4}=\frac{3^{10}\times\left(11+5\right)}{3^9\times16}=\frac{3\times16}{16}=3\)

Chúc bạn học tốt

giúp với mik cần gấp

a.

\(\frac{2^{10}\times13+2^{10}\times65}{2^8\times104}=\frac{2^{10}\times\left(13+65\right)}{2^8\times104}=\frac{2^2\times78}{104}=\frac{4\times78}{104}=\frac{312}{104}=3\)

b.

\(\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(65\times111-15\times37\times13\right)\)

\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(7215-7215\right)\)

\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times0\)

= 0

a. 3

b. 0

\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=2^3=8\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^2.78}{104}=3\)