A = \(\frac{72^3\cdot54^2}{108^4}\)
B = \(\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
K
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16 tháng 9 2017
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)
\(=\frac{2^2.3}{4}+3\)
\(=3+3=6\)
18 tháng 6 2018
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
NH
1
7 tháng 8 2019
\(\frac{3^{10}\times11+3^5\times5}{3^9\times2^4}\)
\(\Rightarrow\)\(\frac{3^5\left(3^5\times11+5\right)}{3^5\times3^4\times2^4}\)
\(\Rightarrow\frac{3^5\times11+5}{\left(3\times2\right)^4}\)
\(\Rightarrow\frac{3^5\times11+5}{6^4}\)
\(\Rightarrow\frac{3^5\times11}{6^4}+\frac{5}{6^4}\)
\(\Rightarrow\frac{3^5\times11}{2^4\times3^4}+\frac{5}{6^4}\)
\(\Rightarrow\frac{3\times11}{16}+\frac{5}{1295}\)
\(\Rightarrow\frac{33}{16}+\frac{5}{1295}\)
\(\Rightarrow\frac{8563}{4144}\)
27 tháng 1 2019
b)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)
27 tháng 1 2019
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) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)
\(=3+3=6\)
N
12 tháng 3 2019
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: \(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2011.2013}\)
\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\)
\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\frac{1}{2}A=1-\frac{1}{2013}\)
\(A=2.\frac{2012}{2013}=\frac{4024}{2013}\)
DL
2
G
23 tháng 9 2020
\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)
\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)
\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)
\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)
\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)
\(=\frac{3}{2^2}=\frac{3}{4}\)
23 tháng 9 2020
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
9 tháng 5 2019
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot10-x=10\)10
\(\left(1-\frac{1}{10}\right)\cdot10-10=x\)
\(x=10\cdot\left(1-\frac{1}{10}-1\right)\)
\(x=10\cdot-\frac{1}{10}=-1\)
9 tháng 5 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right).10-x=10\)
\(\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right].10-x=10\)
\(\left[1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\right].10-x=10\)
\(\left[1-\frac{1}{10}\right].10-x=10\)
\(\frac{9}{10}.10-x=10\)
\(9-x=10\)
\(x=9-10\)
\(x=-1\)
~ Hok tốt ~
\(A=\frac{72^3\times54^2}{108^4}=\frac{\left(2^3\times3^2\right)^3\times\left(2\times3^3\right)^2}{\left(2^2\times3^3\right)^4}=\frac{2^9\times3^6\times2^2\times3^6}{2^8\times3^{12}}=2^3=8\)
\(B=\frac{3^{10}\times11+3^{10}\times5}{3^9\times2^4}=\frac{3^{10}\times\left(11+5\right)}{3^9\times16}=\frac{3\times16}{16}=3\)
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