tim cac so nguyen x,y
a) xy - 2x + y = 5
b) x^2 - xy + x - y = 6
c) (x + 2)^2 + 2(y - 3)^2 < 4
tim cac so nguyen x,y
a) xy-2x+y=5
b) x^2-xy+x-y=6
c) (x+2)^2+2(y-3)^2<4
tim cac so nguyen x,y sao cho:
a) xy-2x+y=5
b) x2-xy+x-y=6
c) (x+2)2+2(y-3)2 < 4
a) xy - 2x + y = 5
10x + y - 2x + y = 5
8x + 2y = 5
8x = 5 - y - y
tim cac so nguyen x,y
a) x^2 - xy + x - y = 6
b) (x+ 2)^2 + 2(y - 3)^2 <4
\(x^2-xy+x-y=\) \(6\)
\(x\left(x-y\right)+\left(x-y\right)\)\(=6\)
\(\left(x+1\right)\left(x-y\right)\) \(=6\)
vì \(x,y\) thuộc Z
=> \(x+1,x-y\)thuộc ước của \(6\)
làm nốt nha
Bai 1 : Tim cac so nguyen x , y biet :
a, x +xy + y = 9
b, xy - 2x -3y = 5
c, xy - 2x + 5y = 2
a, x + xy + y = 9
=>xy + x+y+1=10
=>x.(y+1)+(y+1)=10
=>(x+1).(y+1)= 10.1 = 1.10 = 2.5 = 5.2 = (-10).(-1) = (-1).(-10) = (-2).(-5) = (-5).(-2)
ta có bảng các trường hợp sau
x+1 | 1 | 10 | 2 | 5 | -10 | -1 | -2 | -5 |
y+1 | 10 | 1 | 5 | 2 | -1 | -10 | -5 | -2 |
x | 0 | 9 | 1 | 4 | -11 | -2 | -3 | -6 |
y | 9 | 0 | 4 | 1 | -2 | -11 | -6 | -3 |
vậy
bn tich cho mk nha
tim cac so nguyen x,y cua phuong trinh:x^2+y^2-xy=x+y+2
\(x^2+y^2-xy=x+y+2\)
\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y-4=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=6\)
Vì \(\left(x-y\right)^2+\left(y-1\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-1\right)^2\le6\forall x\)
\(\Rightarrow-\sqrt{6}\le x-1\le\sqrt{6}\)
\(\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Từ đó thay vào tìm các giá trị tương ứng của y.
Tim tat cac cac cap so nguyen x,y thoa man a) x^2+5xy+4y^2
b)xy-2x+3y-1
Tim tat cac cac cap so nguyen x,y thoa man a) x^2+5xy+4y^2
b)xy-2x+3y-1
Tim tat cac cac cap so nguyen x,y thoa man a) x^2+5xy+4y^2
b)xy-2x+3y-1
tim tat ca cac so nguyen x,y thoa man x^3+x^2+2-2y=xy