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\(125^5=\left(5^3\right)^5=5^{15}\)
\(25^7=\left(5^2\right)^7=5^{14}\)
\(\Rightarrow125^5>25^7\)
\(3^{54}=\left(3^6\right)^9=729^9\)
\(2^{81}=\left(2^9\right)^9=512^9\)
\(\Rightarrow3^{54}>2^{81}\)
Bài 1:
\(A=7+7^3+7^5+...+7^{1999}\)
\(\Rightarrow A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)
\(\Rightarrow A=\left(7+343\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)
\(\Rightarrow A=350+7^4.350+...+7^{1996}.350\)
\(\Rightarrow A=\left(1+7^4+...+7^{1996}\right).350⋮35\)
\(\Rightarrow A⋮35\left(đpcm\right)\)
b2:
a) \(S=1+3+3^2+...+3^{49}\)
\(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)
\(\Rightarrow S=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)
\(\Rightarrow S=4+3^2.4+...+3^{48}.4\)
\(\Rightarrow S=\left(1+3^2+...+3^{48}\right).4⋮4\)
\(\Rightarrow S⋮4\left(đpcm\right)\)
c) \(S=1+3+3^2+...+3^{49}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{50}\)
\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)
\(\Rightarrow2S=3^{50}-1\)
\(\Rightarrow S=\frac{3^{50}-1}{2}\left(đpcm\right)\)
1,2 : 10 = 0,12
4,6 : 1000 = 0,0046
781,5 : 100 = 7,815
15,4 : 100 = 0,154
45,82 : 10 = 4,582
15632 : 1000 = 15,632
hok tốt nha ^_^
a: \(M=1+7+7^2+\cdots+7^{2019}\)
\(=\left(1+7+7^2+7^3\right)+\left(7^4+7^5+7^6+7^7\right)+\cdots+\left(7^{2016}+7^{2017}+7^{2018}+7^{2019}\right)\)
\(=\left(1+7+7^2+7^3\right)+7^4\left(1+7+7^2+7^3\right)+\cdots+7^{2016}\left(1+7+7^2+7^3\right)\)
\(=400\left(1+7^4+\cdots+7^{2016}\right)\)
=>M⋮10
=>M có chữ số tận cùng là 10
b: \(N=3+3^3+3^5+\cdots+3^{101}\)
\(=3+\left(3^3+3^5\right)+\left(3^7+3^9\right)+\cdots+\left(3^{99}+3^{101}\right)\)
\(=3+3^3\left(1+3^2\right)+3^7\left(1+3^2\right)+\cdots+3^{99}\left(1+3^2\right)\)
\(=3+10\left(3^3+3^7+\cdots+3^{99}\right)\)
=>N chia 10 dư 3
=>N có chữ số tận cùng là 3