CMR: \(\left(1\cdot2\cdot3\cdot..\cdot90-60\right)\) chia hết cho 5
DA
Những câu hỏi liên quan
N=\(\frac{5\cdot\left(2^2\cdot3^2\right)\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot2^{18}}\)
Tinh N
Tính
\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}+7\cdot2^{29}\cdot3^{18}}\)
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)
\(=\frac{5.4-2}{19}=\frac{18}{19}\)
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thực hiện phép tính
A=\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
A =\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\cdot3^4\)
Hãy thực hiện phép tính
tính
A = \(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-\left(2^2\cdot3\right)^{14}9^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
ta có \(\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-\left(2^2.3\right)^{14}.9^{14}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2^{28}.3^{14}.3^8}{2^{28}.3^{18}\left(5.1.1-7.2.1\right)}\)
\(=\frac{2^{28}.3^{18}\left(5.1.3.2^2-1.3^4\right)}{2^{28}.3^{18}\left(5-14\right)}\)
\(=\frac{60-81}{5-14}=\frac{7}{3}\)
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cho Sn= \(\frac{1}{1\cdot2\cdot3\cdot4}\)+ \(\frac{1}{2\cdot3\cdot4\cdot5}\)+ ... + \(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\)
CMR: 18<\(\frac{1}{S_n}\)<=24
rút gọn \(B=\frac{5}{1\cdot2\cdot3}+\frac{5}{2\cdot3\cdot4}+....+\frac{5}{n\cdot\left(n+1\right)\left(n+2\right)}\)
\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)
\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)
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Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
tính bằng cách hợp lí
a) \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
b) \(1\cdot2\cdot3\cdot4\cdot...\cdot8\cdot9-1\cdot2\cdot3\cdot4\cdot...\cdot7\cdot8-1\cdot2\cdot3\cdot...\cdot7\cdot8^2\)
c) \(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}\)
giúp em với ạ
c) \(\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{11\cdot2^{35}-2^{26}}\)
\(=\frac{9\cdot2^4\cdot2^{32}2^{ }}{\left(11-2\right)\cdot2^{35}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{9\cdot2^{35}}\)
\(=\frac{9\cdot1\cdot2^{32}}{9\cdot2^{31}}=\frac{2^{32}}{2^{31}}=2\)
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