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![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a)<=>(n-1)+4 chia hết n-1
=>4 chia hết n-1
=>n-1\(\in\){-1,-2,-4;1,2,4}
=>n\(\in\){0,-1,-3,2,3,5}
b)<=>2(2n+1)+2 chia hết 2n+1
=>4 chia hết 2n+1
=>2n+1\(\in\){-1,-2,-4,1,2,4}
=>n\(\in\){-1;-3;-7;3;5;9}
bài 3 : <=>2y+8+xy+4x-1y-4=11
=>(8-4)+(2y-1y)+xy+4x=11
=>4+1y+x.y+x.4=11
=>1y+x.(x+y)=11-4
=>y+x.x+y=8
=>(x+y)^2=8
=>x+y=3
=>x và y là các số có tổng =3 ( bn tự liệt kê nhé )
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{1;0;2;-1\right\}\)
c: \(\Leftrightarrow n+1\in\left\{1;-1\right\}\)
hay \(n\in\left\{0;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4n+1}{2n+3}=\frac{4n+6-5}{2n+3}=\frac{2\left(2n+3\right)-5}{2n+3}=2-\frac{5}{2n+3}\)
Để \(2-\frac{5}{2n+3}\) là số nguyên <=> \(\frac{5}{2n+3}\) là số nguyên
=> 2n + 3 thuộc Ư(5) = { - 5; - 1; 1; 5 }
=> 2n + 3 = { - 5; - 1; 1; 5 }
=> n = { - 4; - 2; - 1 ; 1 }
\(4n-5⋮2n-1\)
\(\Rightarrow4n-2-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)\)
\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2n-1=1\Rightarrow2n=2\Rightarrow n=1\\2n-1=-1\Rightarrow2n=0\Rightarrow n=0\\2n-1=3\Rightarrow2n=4\Rightarrow n=2\\2n-1=-3\Rightarrow2n=-2\Rightarrow n=-1\end{matrix}\right.\)
2) \(A=\dfrac{9n+7}{3n+4}=\dfrac{9n+12-5}{3n+4}=\dfrac{9n+12}{3n+4}-\dfrac{5}{3n+4}=\dfrac{3\left(3n+4\right)}{3n+4}-\dfrac{5}{3n+4}=3-\dfrac{5}{3n+4}\)
\(\Rightarrow5⋮3n+4\)
\(\Rightarrow3n+4\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(MIN_A\Rightarrow MAX_{3n+4}\)
\(\Rightarrow3n+4=-1\Rightarrow3n=-5\Rightarrow n=-\dfrac{5}{3}\)
Tương tự
\(\Rightarrow\left[{}\begin{matrix}3n+4=1\Rightarrow3n=-3\Rightarrow n=-1\\3n+4=-1\Rightarrow3n=-5\Rightarrow n=-\dfrac{5}{3}\\3n+4=5\Rightarrow3n=1\Rightarrow n=\dfrac{1}{3}\\3n+4=-5\Rightarrow3n=-9\Rightarrow n=-3\end{matrix}\right.\)
b) \(MIN_A\Rightarrow A\in Z^-\Rightarrow3n+4\in Z^-\)
@Azue