cho phan so A=5n-19/n-4 tim n thuoc z de A thuoc z
Cho phan so: P = \(\frac{2n+7}{5n+2}\)(n thuoc Z)
a)Tim n thuoc z de P la phan so chua toi gian.
b)Tim n thuoc Z sao cho P la phan so toi gian.
cho A=n+3\n-2 (n thuoc Z)
a) tim n de A la phan so
b) tim n de A thuoc Z
c) tim n biet A=-4
Tim n thuoc Z de phan so sau rut gon duoc:
4n+5/5n+4
Nho ghi cach lam nua nhe
Cho phan so A=n+10/2n-8 tim so n de A thuoc Z
Cho phan so A = 2n + 9/n + 3. Tim n de A thuoc Z
Cho phan so A= n-2/n+3
Tim n de
a) a la phan so
b) A thuoc Z
a: Để A là phân số thì n+3<>0
hay n<>-3
b: Để A là số nguyên thì \(n-2⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(-5\right)\)
\(\Leftrightarrow n+3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-2;-4;2;-8\right\}\)
Cho phan so A= n-2/n+3
Tim n de
a) a la phan so
b) A thuoc Z
a) Để A là phân số \(\Leftrightarrow n+3\ne0\Leftrightarrow n\ne-3\)
b) \(A=\frac{n-2}{n+3}=\frac{\left(n+3\right)-5}{n+3}=1-\frac{5}{n+3}\)
để A nguyên thì \(n+3\inƯ\left(5\right)\)
Mà: Ư(5)={1;-1;5;-5}
Ta có bảng sau:
n+3 | 1 | -1 | 5 | -5 |
n | -2 | -4 | 2 | -8 |
Vậy n={-8;-4;-2;2}
a) Để A là phân số thì \(n+3\ne0\Leftrightarrow n\ne-3\)
b) Ta có:
\(A=\frac{n-2}{n+3}=\frac{n+3-5}{n+3}=\frac{n+3}{n+3}-\frac{5}{n+3}=1-\frac{5}{n+3}\)
Để \(A\in Z\Leftrightarrow\frac{5}{n+3}\in Z\)
\(\Rightarrow n+3\inƯ\left(5\right)\)
\(\Rightarrow n+3\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow n\in\left\{-2;-4;2;-8\right\}\)
Vậy \(n\in\left\{-2;-4;2;-8\right\}\)
1)cho phan so A=\(\frac{n+1}{n-2}\)
a tim n thuoc Z de A co gia tri nguyen
b tim n thuoc Z de A co gia tri lon nhat
\(\frac{n+1}{n-2}=\frac{\left(n-2\right)+3}{n-2}=\frac{n-2}{n-2}+\frac{3}{n-2}=1+\frac{3}{n-2}\)
Để \(\frac{3}{n-2}\in Z\) <=> 3 ⋮ n - 2 => n - 2 ∈ Ư ( 3 ) = { - 3 ; - 1 ; 1 ; 3 }
=> n ∈ { - 1 ; 1 ; 3 ; 5 }
cho phan so A = n - 5 / n+1 ( n thuoc Z , n khac -1)
a. Tim n de A toi gian
de A toi gian thi
n-5 chia het cho n+1
(n+1)-6 chia het cho n+1
vi n+1 chia het cho n+1
nen -6 chia het cho n+1
=> n+1 thuoc U(-6)=(1;-1;2;-2;3;-3;6;-6)
n+1 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
n | 0 | -2 | 1 | -3 | 2 | -4 | 5 | -7 |
=> n thuoc (0;-2;1;-1;-3;2;-4;5;-7)