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Bài 1:
\(A=2+2^2+2^3+...+2^{2003}+2^{2004}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+2^4+...+2^{2002}\right)⋮7\)
Bài 2:
\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\cdot\left(2+2^5+...+2^{57}\right)⋮15\)
Bài 3:
\(A=1+3+3^2+3^3+...+3^{1990}+3^{1991}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1989}+3^{1990}+3^{1991}\right)\)
\(=13+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
Bài 4:
\(A=4+4^2+4^3+4^4+...+4^{23}+4^{24}\)
\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{23}+4^{24}\right)\)
\(=\left(4+4^2\right)+4^2\left(4+4^2\right)+...+4^{22}\left(4+4^2\right)\)
\(=20\left(1+4^2+...+4^{22}\right)⋮20\)
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a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
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Ư(240)={1;2;3;4;5;6;8;10;12;15;16;20;24;30;40;48;60;80;120;240}
Trong các số này thì các số là bội của 24 là:
24;48;120;240
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
Ta có: 6n + 4 \(⋮\)2n + 1 (n \(\inℤ\))
=> 6n + 4 - 3(2n + 1) \(⋮\)2n + 1
=> 1 \(⋮\)2n + 1
=> 2n + 1 \(\in\)Ư (1)
Ư (1) = {1; -1}
2n + 1 = 1 hay -1
2n = 1 - 1 hay -1 - 1
2n = 0 hay -2
n = 0 : 2 hay -2 : 2
n = 0 hay -1
Vậy n = 0 hay -1
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a) \(\left(0,5\right)^{12}:\left(0,5\right)^{10}=\left(0,5\right)^{12-10}=\left(0,5\right)^2\)
b) \(\sqrt{36}=\pm6\)
c)\(\left(0,75\right)^{22}:\left(0,75\right)^{12}=\left(0,75\right)^{22-12}=\left(0,75\right)^{10}\)
d) \(\sqrt{49}=\pm7\)
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\(100:\left\{2\cdot\left[30-\left(12+7\right)\right]\right\}\)
\(=100:\left[2\cdot\left(30-19\right)\right]\)
\(=100:\left(2\cdot11\right)\)
\(=100:22\)
\(=\dfrac{100}{22}\)
\(=\dfrac{50}{11}\)
\(A=3+3^2+...+3^{2005}\)
\(\Rightarrow3A=3^2+3^3+...+3^{2006}\)
\(\Rightarrow3A-A=3^{2006}-3\)
\(\Rightarrow2A=3^{2006}-3\)
\(\Rightarrow2A+3=3^{2006}\) là 1 lũy thừa của 3 (đpcm)
4.
\(B=1+1+2+2^2+2^3+...+2^{100}\)
\(2B=2+2+2^2+...+2^{101}\)
\(\Rightarrow2B-B=2+2^{101}-\left(1+1\right)=2^{101}\)
\(\Rightarrow B=2^{101}\) là 1 lũy thừa của 2 (đpcm)