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![](https://rs.olm.vn/images/avt/0.png?1311)
a,
Ta có: 4n-5 chia hết cho 2n-1
=>4n-2-3 chia hết cho 2n-1
=>2.(2n-1)-3 chia hết cho 2n-1
=>3 chia hết cho 2n-1
=>2n-1=Ư(3)=(-1,-3,1,3)
=>2n=(0,-2,2,4)
=>n=(0,-1,1,2)
Vậy n=0,-1,1,2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{1;0;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n-1\in\left\{-1;1;3\right\}\)
hay \(n\in\left\{0;1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)
b) \(\left(n-1\right)\inƯ\left(28\right)=\left\{1;2;4;7;14;28\right\}\)
\(\Rightarrow n\in\left\{2;3;5;8;15;29\right\}\)
c) \(\left(2n+1\right)\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
\(\Rightarrow2n\in\left\{0;1;2;5;8;17\right\}\)
Mà \(n\in N\)
\(\Rightarrow n\in\left\{0;1;4\right\}\)
d) \(n\left(n+2\right)=8\)
\(\Leftrightarrow n^2+2n-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(n+6\right)⋮\left(n+1\right)\Rightarrow\left(n+1\right)+5⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;4\right\}\)
b) \(\left(4n+9\right)⋮\left(2n+1\right)\Rightarrow2\left(2n+1\right)+7⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(n\in\left\{1;7\right\}\)
b: \(n-1\in\left\{-1;1;7\right\}\)
hay \(n\in\left\{0;2;8\right\}\)
c: \(2n-1\in\left\{-1;1;7\right\}\)
\(\Leftrightarrow2n\in\left\{0;2;8\right\}\)
hay \(n\in\left\{0;1;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a.`
`3^n = 27` phải k c?
`3^n = 27`
`=> 3^n = 3^3`
`=> n=3`
Vậy, `n=3`
TH2 (đề):
`3n = 27`
`=> n = 27 \div 3`
`=> n=9`
Vậy, `n=9`
`b.`
TH1:
`5^n = 625`
`=> 5^n = 5^4`
`=> n = 4`
Vậy, `n=4`
TH2:
`5n = 625`
`=> n = 625 \div 5`
`=> n = 125`
Vậy, `n=125`
5n = 625
=> 5n = 54
=> n = 4
n2 = 169
=> n2 = 132
=> n = 13
62n = 1296
=> 62n = 64
=> 2n = 4
=> n = 2
14n = 149 : 2744
=> 14n = 149 : 143
=> 14n = 146
=> n = 6
Study well
a)Ta có 5n=625
=>5n=54
=>n=4
b)Ta có n2=169
=>n2=132
=>n=±13
c)Ta có 62n=1296
=>62n=64
=>2n=4
=>n=2
d) Ta có 14n=149:2744
=>14n=149:143
=>14n=146
=>n=6
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