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câu a là vô tận
b)Vì \(\frac{3n+4}{n-2}\in Z\Rightarrow3n+4⋮n-2\Rightarrow3n-6+10⋮n-2\)
\(\Rightarrow10⋮n+2\Rightarrow n+2\inƯ\left(10\right)\)
đến đó bạn tự làm nhé
a, Gọi ƯCLN 2n + 5 ; n + 3 = d \(\left(d\inℕ^∗\right)\)
Ta có : \(2n+5⋮d\)(1)
\(n+3⋮d\Rightarrow2n+6⋮d\)(2)
Lấy (2) - (1) ta được : \(2n+6-2n-5⋮d\Rightarrow1⋮d\Rightarrow d=1\)
b, Để \(B=\frac{2n}{n+3}+\frac{5}{n+3}=\frac{2n+5}{n+3}\)nhận giá trị nguyên khi
\(2n+5⋮n+3\Leftrightarrow2\left(n+3\right)-1⋮n+3\)
\(\Rightarrow n+3\inƯ\left(1\right)=\left\{\pm1\right\}\)
n + 3 | 1 | -1 |
n | -2 | -4 |
DKXD cua phan thuc \(n\ne-9\)
\(\frac{7n-1}{n+9}=\frac{7n+63-64}{n+9}=\frac{7\left(n+9\right)-64}{n+9}=\frac{7\left(n+9\right)}{n+9}-\frac{64}{n+9}\)\(=7-\frac{64}{n+9}\)
De phan thuc dat gia tri nguyen => \(\frac{64}{n+9}\)nguyen
<=> \(64⋮n+9\)<=> \(n+9\in U\left(64\right)\)
<=> \(n+9\in\left\{-64;-32;-16;-8;-4;-2;-1;1;2;4;8;16;32;64\right\}\)
=> \(n\in\left\{-73;-41;-25;-17;-13;-11;-10;-7;-5;-1;7;23;55\right\}\)
Ta có: \(A=\frac{2n-1}{n+3}=2-\frac{7}{n+3}\)
Để A nguyên thì \(7\)\(⋮\)\(n+3\)
\(\Rightarrow\)\(n+3\)\(\inƯ\left(7\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\)\(n\)\(=\left\{-10;-4;-2;4\right\}\)
\(A=\frac{2n-1}{n+3}\) có giá trị nguyên
\(\Leftrightarrow2n-1⋮n+3\)
\(\Rightarrow\left(2n+6\right)-6-1⋮n+3\)
\(\Rightarrow2\left(n+3\right)-7⋮n+3\)
có \(2\left(n+3\right)⋮n+3\)
\(\Rightarrow-7⋮n+3\)
\(\Rightarrow n+3\inƯ\left(-7\right)\)
\(n\in Z\Rightarrow n+3\in Z\)
\(\Rightarrow n+3\in\left\{-1;-7;1;7\right\}\)
\(\Rightarrow n\in\left\{-4;-10;-2;4\right\}\)
\(\frac{8n-3}{2n+1}=\frac{13}{5}\)
\(\Rightarrow\left(8n-3\right)\cdot5=\left(2n+1\right)\cdot13\)
\(\Rightarrow40n-15=26n+13\)
\(\Rightarrow40n-26n=13+15\)
\(\Rightarrow14n=28\)
\(\Rightarrow n=28\div2\)
\(\Rightarrow n=14\)
ta có : 8n-3/2n+1=13/5
(8n-3).5=(2n+1).13
40n-15=26n+13
40n-26n=15+23
14n=28
suy ra n=28:14=2
vậy n=2
a, \(A=\frac{7}{n-3}\)
Để \(\frac{7}{n-3}\in Z\)thì \(7⋮n-3\Leftrightarrow n-3\inƯ\left(7\right)=\left\{\text{±}1;\text{±}7\right\}\)
Ta có bảng sau:
n - 3 | -1 | -7 | 1 | 7 |
n | 2 | -4 | 4 | 10 |
Vậy \(n\in\left\{-4;2;4;10\right\}\)để\(\frac{7}{n-3}\in Z\)
b,\(B=\frac{13}{2n-5}\)
Để \(\frac{13}{2n-5}\in Z\)thì \(13⋮2n-5\Leftrightarrow2n-5\inƯ\left(13\right)=\left\{\text{±}1;\text{±}13\right\}\)
Ta có bảng sau:
2n - 5 | -1 | -13 | 1 | 13 |
2n | 4 | -8 | 6 | 18 |
n | 2 | -4 | 3 | 9 |
Vậy \(n\in\left\{-4;2;3;9\right\}\)để\(\frac{13}{2n-5}\in Z\)
c, \(C=\frac{-6}{3n+2}\)
Để \(\frac{-6}{3n+2}\in Z\)thì \(-6⋮3n+2\Leftrightarrow3n+2\inƯ\left(-6\right)=\left\{\text{±}1;\text{±}2;\text{±}3;\text{±}6\right\}\)
Ta có bảng sau:
3n + 2 | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 |
3n | -3 | -4 | -5 | -8 | -1 | 0 | 1 | 4 |
n | -1 | \(\frac{-4}{3}\) | \(\frac{-5}{3}\) | \(\frac{-8}{3}\) | \(\frac{-1}{3}\) | 0 | \(\frac{1}{3}\) | \(\frac{4}{3}\) |
Vậy \(n\in\left\{\frac{-8}{3};\frac{-5}{3};\frac{-4}{3};\frac{-1}{3};-1;0;\frac{1}{3};\frac{4}{3}\right\}\)để \(\frac{-6}{3n+2}\in Z\)
mà \(n\in Z\)
Vậy \(n\in\left\{-1;0\right\}\)để\(\frac{-6}{3n+2}\in Z\)
a,Để \(A\in Z\)
\(\Rightarrow\)\(\frac{7}{n-3}\in Z\)
\(\Rightarrow\)n-3\(\in\)Ư(7)
n-3 \(\in\){1;-1;7;-7}
n\(\in\){4;2;10;-4}
Vậy n\(\in\){4;2;10;-4}
b,Để \(B\in Z\)
\(\Rightarrow\frac{13}{2n-5}\in Z\)
\(\Rightarrow\)2n-5\(\in\)Ư(13)
2n-5\(\in\){1;-1;13;-13}
2n\(\in\){6;4;18;-8}
n\(\in\){3;2;9;-4}
Vậy n\(\in\){3;2;9;-4}
c,Để \(C\in Z\)
\(\Rightarrow\frac{-6}{3n+2}\in Z\)
\(\Rightarrow\)3n+2\(\in\)Ư(-6)
3n+2\(\in\){1;-1;2;-2;3;-3;6;-6}
n\(\in\){-1;0}
Vậy n \(\in\){-1;0}
Ta có : \(\frac{n-3}{n+4}=\frac{n+4-7}{n+4}=\frac{n+4}{n+4}-\frac{7}{n+4}=1-\frac{7}{n+4}\)
Để \(\frac{n-3}{n+4}\in Z\) thì 7 chia hết cho n + 4
=> n + 4 thuộc Ư(7) = {-7;-11;7}
Ta có bảng :
n + 4 | -7 | -1 | 1 | 7 |
n | -11 | -5 | -3 | 3 |
\(\frac{8n-9}{2n+5}=\frac{8n+20-20-9}{2n+5}=\frac{8n+20-29}{2n+5}=\frac{8n+20}{2n+5}+\frac{-29}{2n+5}\)
\(Ư\left(-29\right)=\left(-29;-1;1;29\right)\)
\(2n+5=-29\) \(n=-17\)
\(2n+5=-1\) \(n=-3\)
\(2n+5=1\) \(n=-2\)
\(2n+5=29\) \(n=12\)
\(n=\left(-17;-3;-2;12\right)\)