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\(2n+9⋮3n+1\)
\(\Rightarrow3\left(2n+9\right)⋮3n+1\)
\(\Rightarrow2\left(3n+1\right)+25⋮3n+1\)
\(\Rightarrow25⋮3n+1\)
\(\Rightarrow3n+1\in\left\{5,25,1,-5,-25,-1\right\}\)
\(n\in\left\{8,0\right\}\)
\(5n+2⋮9-2n\)
\(\Rightarrow2\left(5n+2\right)⋮9-2n\)
\(\Rightarrow-5\left(9-2n\right)-41⋮9-2n\)
\(41⋮9-2n\)
\(\Rightarrow9-2n\in\left\{41,-41,1,-1\right\}\)
\(\Rightarrow n\in\left\{-16,25,4,-5\right\}\)
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a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
n-3 | 1 | -1 | 2 | -2 | 4 | -4 |
n | 4 | 2 | 5 | 1 | 7 | -1 |
a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)
\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)
\(A=\dfrac{n+1}{n-3}\)
\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)
\(A=1+\dfrac{4}{n-3}\)
Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
n-3=1 --> n=4
n-3=-1 --> n=2
n-3=2 --> n=5
n-3=-2 --> n=1
n-3=4 --> n=7
n-3=-4 --> n=-1
Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên
b.hemm bt lèm:vv
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\(------huongdan-----\)
\(Taco:\)
\(\left(3n-2n\right)⋮n+1\Leftrightarrow n⋮n+1\Leftrightarrow\left(n+1\right)-n⋮n+1\Leftrightarrow1⋮n+1\)
\(\Leftrightarrow n+1\in\left\{-1;1\right\}\Leftrightarrow n\in\left\{-2;0\right\}\)
\(b,2n-4⋮n+2\Leftrightarrow2n+4-2n+4⋮2n+4\Leftrightarrow8⋮2n+4\)
dễ thấy: 2n+4 chẵn => 2n+4 là ước chẵn của 8
\(\Rightarrow2n+4\in\left\{2;4;8;-2;-4;-8\right\}\Rightarrow2n\in\left\{-2;0;4;-6;-8;-12\right\}\)
\(\Rightarrow n\in\left\{-1;0;2;-3;-4;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n^2-3n+9\)chia het cho \(n-2\)
\(\Leftrightarrow\)\(n^2-2n-n-2+11\)chia het cho \(n-2\)
\(\Leftrightarrow\)\(\left(n-2\right)\left(n+1\right)+11\)chia het cho \(n-2\)
\(\Leftrightarrow\)11 chia het cho \(n-2\)
\(\Rightarrow\)\(n-2\in U\left(11\right)\)\(\Rightarrow\)\(n-2\in\left\{-11;-1;1;11\right\}\)
\(\Rightarrow\)\(n\in\left\{-9;1;3;13\right\}\)
b) 2n-1 chia hết cho n-2
\(\Rightarrow2n-2+3\) chia hết cho\(n-2\)
\(\Rightarrow3\)chia hết cho \(n-2\)
\(\Rightarrow n-2\in U\left(3\right)\)\(\Rightarrow n-2\in\left\{-3;-1;1;3\right\}\)\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)
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1) Gọi \(d=ƯCLN\left(2n+1;3n+2\right)\)
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}\)
\(\Rightarrow2\left(3n+2\right)-3\left(2n+1\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\Rightarrow2n+1\)và\(3n+2\)là nguyên tố cùng nhau
\(\Rightarrow\frac{2n+1}{3n+2}\)là phân số tối giản\(\left(đpcm\right)\)
câu 1 :
gọi d = ƯCLN ( 2n + 1; 3n +2 )
=> 2n + 1 chia hết cho d => 3 ( 2n +1 ) chia hết cho d
3n + 2 chia hết cho d => 2 ( 3n + 2 ) chia hết cho d
ta có : 3 ( 3n + 2 ) - [ 2 ( 2n + 21) ] hay 6n + 4 - [ 6n + 3 ] chia hết cho d
=> 1 chia hết cho d -> 2n +1 và 3n + 2 là hai số nguyên tố cùng nhau
=> \(\frac{2n+1}{3n+2}\) là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vì 3\(⋮\)n
=> n\(\in\)Ư(3)={ 1; 3 }
Vậy, n=1 hoặc n=3
2.(3n-4)=6n-8 chia hết (2n-1)
6n-8=3(2n-1)-5 chia hết (2n-1)
2n-1 chia hết cho (2n-1) hiển nhiên
=> 5 phải chia hết cho (2n-1)
2n-1 = ước (5) =(-5,-1,1,5)
2n=(-4,0,2,6)
n={-2,0,1,3}