(2x+1)(y-2)=3
=> 2x+1; y-3 thuộc Ư(3)={-1,-3,1,3}
Ta có bảng
2x+1 | -1 | -3 | 1 | 3 |
y-3 | 3 | 1 | -3 | 1 |
x | -1 | -2 | 0 | 1 |
y | 6 | 4 | 0 | 4 |
Vậy ta có các cặp số x,y thõa mãn (-1,6);(-2,4);(0,0);(1,4)
(2x+1)(y-2)=3
=> 2x+1; y-3 thuộc Ư(3)={-1,-3,1,3}
Ta có bảng
2x+1 | -1 | -3 | 1 | 3 |
y-3 | 3 | 1 | -3 | 1 |
x | -1 | -2 | 0 | 1 |
y | 6 | 4 | 0 | 4 |
Vậy ta có các cặp số x,y thõa mãn (-1,6);(-2,4);(0,0);(1,4)
tim cac so nguyen y , x biet
(2x + 1)(y - 4) = 6
Tim cac cap so nguyen x;y biet : (2x-1)*(y-4)=-13
1,tim cac so nguyen x,y biet: -2/x=y/3 va x<0<y
2, Tim cac so nguyen x,y biet:x-3/y-2=3/2 va x-y=4
Tim cac gia tri nguyen cua x biet
a, Y=5x+9/ x+3
b, Tim cac cap so nguyen x , y thoa mai he thuc :
(2x-1).(y+4)
tim cac so nguyen x , y biet
c) (2x + 1)(y - 4)
d) x + y = 3 va x - y = 15
Bai 1 : Tim cac so nguyen x , y biet :
a, xy - 2x -3y = 5
b, xy - 2x + 5y = 2
Bai 1 : Tim cac so nguyen x , y biet :
a, x +xy + y = 9
b, xy - 2x -3y = 5
c, xy - 2x + 5y = 2
tim cac so nguyen x,y,z biet y^2=y-1; x^2=x-1; z^2=z-1
Tim x biet:
(2x-1)5=(2x-1)7
Tim cac so nguyen to x , y thoa man:
(x-2)2.(y-3)2=4
trinh bay cach lam nhe