tim so nguyen a de a^2 + a + 3 / a+1 la so nguyen
NT
Những câu hỏi liên quan
Cho A=n+3/n+2 voi n€z
a)tim dieu kien cua so nguyen n de A la phan so
b)tinh gia tri cua phan so A khi n=1;n=-1
c)tim so nguyen n de phan so A co gia tri la so nguyen
Cho A= n-1/n+4
a) Tim n nguyen de A la mot phan so
b) Tim n nguyen de A la mot so nguyen
Cho A = n-1/n+4
a,tim n de A la so nguyen to
b,tim n nguyen de A la mot so nguyen
Cho bieu thuc A = 3/n-2
a) Tim cac so nguyen n de bieu thuc A la phan so
b) Tim cac so nguyen n de A la mot so nguyen
Cac giup minh nhe minh dang can gap.
cho A=n-1/n-4
tim n nguyen de A la mot phan so
tim n nguyen de n la mot so nguyen
A=n-1/n+2
a, Voi gia tri nao thi A la phan so
b, Tim so nguyen n de A=0
c,tim so nguyen n de A co gia tri nguyen
bai 1:tim so nguyen x sao cho gia cua cac phan so la so nguyen
a)x+3/x-2
b)x2+3x-2/x+2
chu y:dau / la dau chia trong phan so
bai 2:cho A=2n+1/n-2 voi n la so nguyen
a)tim n de a la phan so
b)tim n de A nguyen
c)tinh gia tri cua A biet:n=2;1;-2;-1
giup minh voi minh dang can.Ai dung minh tick cho
a) Tim so nguyen a de a2 + a + 3 / a+1 la so nguyen.o cho x-2xy+y=0
b)Tim so nguyen x,y sao cho x-2xy+y=0.
Oái gặp bn trùng tên nè!
a) Để phân số \(\dfrac{a^2+a+3}{a+1}\) là số nguyên thì :
\(a^2+a+3⋮a+1\)
Mà \(a+1⋮a+1\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+a+3⋮a+1\\a^2+a⋮a+1\end{matrix}\right.\)
\(\Rightarrow3⋮a+1\)
Vì \(a\in Z\Rightarrow a+1\in Z;a+1\inƯ\left(3\right)\)
Ta có bảng :
| \(a+1\) | \(1\) | \(3\) | \(-1\) | \(-3\) |
| \(a\) | \(0\) | \(2\) | \(-2\) | \(-4\) |
| \(Đk\) \(a\in Z\) | TM | TM | TM | TM |
Vậy \(a\in\left\{0;2;-2;-4\right\}\) là giá trị cần tìm
b) Ta có :
\(x-2xy+y=0\)
\(\Rightarrow2x-4xy-2y=0\)
\(\Rightarrow\left(2x-4xy\right)+2y-1=0-1\)
\(\Rightarrow\left(2x-4xy\right)-\left(1-2y\right)=-1\)
\(\Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Rightarrow\left(1-2y\right)\left(2x-1\right)=-1\)
Vì \(x,y\in Z\Rightarrow1-2y;2x-1\in Z,1-2y;2x-1\inƯ\left(-1\right)\)
Ta có bảng :
| \(x\) | \(2x-1\) | \(1-2y\) | \(y\) | \(Đk\) \(x,y\in Z\) |
| \(0\) | \(-1\) | \(1\) | \(0\) | TM |
| \(1\) | \(1\) | \(-1\) | \(1\) | TM |
Vậy cặp giá trị \(\left(x,y\right)\) cần tìm là :
\(\left(0,0\right);\left(1,1\right)\)
Đúng 0
Bình luận (3)
b) \(x-2xy+y=0\)
\(\Rightarrow x-\left(2xy-y\right)=0\)
\(\Rightarrow x-y\left(2x-1\right)=0\)
\(\Rightarrow2x-2y\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)-2y\left(2x-1\right)=0-1=-1\)
\(\Rightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
Ta có:
TH1: \(\left\{{}\begin{matrix}2x-1=1\\1-2y=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
TH2:\(\left\{{}\begin{matrix}2x-1=-1\\1-2y=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy...................
Đúng 0
Bình luận (0)
cho A=n-1/n+4
a) tim n nguyen de A la mot phan so
b) tim n nguyen de A la mot so nguyen