Question

[thảo nguyễn thanh]

Bài 8 : Tìm n \(\in\)z để các phân số có giá trị nguyên:

a, A = \(\frac{7}{n-3}\) 

b, B = \(\frac{13}{2n-5}\)

c, C = \(\frac{-6}{3n+2}\)

Answer 1

[Thái Sơn Phạm]

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\(\frac{-6}{3n+2}\in Z\)

Answer 2

[Hoàng Văn Dũng]

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}