Question

Aflati perechile (x,y) de numere intregi cu proprietatea ca:
2(x^2+2y^2)-8(x-y)+1<0
P.S: Este mai mic sau egal cu 0

Answer 1

[tex]2x^2+4y^2-8x+8y+1=2(x^2-4x)+4(y^2+2y)+1=\\ 2(x^2-4x+4)-8+4(y^2+2y+1)-4+1=\\ 2(x-2)^2+4(y+1)^2-11\leq0\\ 2(x-2)^2+4(y+1)^2\leq11\\ (x-2)^2+2(y+1)^2\leq\frac{11}{2}=5.5\\ (x-2)^2+2(y+1)^2\in\{0,1,2,3,4,5\}\\ x-2=0,\ y+1=0\Rightarrow x=2,y=-1\\ x-2=1\ sau\ x-2=-1,y+1=0\Rightarrow x=3,y=-1sau\ x=1,y=-1\\ x-2=0,y+1=1\ sau\ y+1=-1 \Rightarrow x=2, y=0\ sau x=2,y=-2 \\ \text{...samd...se iau toate cazurile... sunt destul de multe } [/tex]