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

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

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

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

\(=3+3=6\)

B=3^10.11+3^10.5/3^9.2^4

  = 3^10( 11+5)/3^9.16

  = 3^10.16/3^9.16

  = 3^10/3^9

  = 3

Vậy B = 3 (1)

C = 2^10.13+2^10.65/2^8.104

   = 2^10(13+65)/2^8.2^2.26

   = 2^10.78/2^10.26

   = 78/26

   = 3

Vậy C = 3 (2)

Từ (1) v (2) suy ra B=C

\(\frac{3^{10}\times11+3^5\times5}{3^9\times2^4}\)

\(\Rightarrow\)\(\frac{3^5\left(3^5\times11+5\right)}{3^5\times3^4\times2^4}\)

\(\Rightarrow\frac{3^5\times11+5}{\left(3\times2\right)^4}\)

\(\Rightarrow\frac{3^5\times11+5}{6^4}\)

\(\Rightarrow\frac{3^5\times11}{6^4}+\frac{5}{6^4}\)

\(\Rightarrow\frac{3^5\times11}{2^4\times3^4}+\frac{5}{6^4}\)

\(\Rightarrow\frac{3\times11}{16}+\frac{5}{1295}\)

\(\Rightarrow\frac{33}{16}+\frac{5}{1295}\)

\(\Rightarrow\frac{8563}{4144}\)

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

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

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

\(=3+3=6\)

Bài 5:

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

\(A=1-\frac{1}{50}< 1\)

Vậy A<1.

Bài 4: Bn ghi nhầm đề rồi.

Đề đúng: \(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2011.2013}\)

\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\)

\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(\frac{1}{2}A=1-\frac{1}{2013}\)

\(A=2.\frac{2012}{2013}=\frac{4024}{2013}\)

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)

\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)

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

Bài làm :

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{\left(2.3\right)^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{2^{10}.3^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{2^8.3^9.\left(2^2.3-5\right)}{3^8.2^{10}.\left(3+2^2\right)}\)

\(=\frac{3.7}{2^2.7}\)

\(=\frac{3}{4}\)

Học tốt

\(\frac{16}{11},-\frac{5}{9},\frac{10}{539}\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot10-x=10\)10

\(\left(1-\frac{1}{10}\right)\cdot10-10=x\)

\(x=10\cdot\left(1-\frac{1}{10}-1\right)\)

\(x=10\cdot-\frac{1}{10}=-1\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right).10-x=10\)

\(\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right].10-x=10\)

\(\left[1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\right].10-x=10\)

\(\left[1-\frac{1}{10}\right].10-x=10\)

\(\frac{9}{10}.10-x=10\)

\(9-x=10\)

\(x=9-10\)

\(x=-1\)

~ Hok tốt ~

\(=>\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot10-x=10\)

\(=>\left(1-\frac{1}{10}\right)\cdot10-x=10\)

\(=>\frac{9}{10}\cdot10-x=10\)

\(=>x=9-10=-1\)