Question

Tìm x,y nguyên biết : x+xy-x²+y=1

Answer 1

x+xy-\(x^2\)+y=1

<=>xy+y=\(x^2\)-x+1(*)

.Nếu x+1=0=>x=-1=>0.y=3->vô lí

Nếu x+1\(\ne\)0=>y=\(\frac{x^2-x+1}{x+1}=\frac{x^2+x-2x+1}{x+1}=\frac{x^2+x}{x+1}+\frac{-2x+1}{x+1}\)=\(\frac{x\left(x+1\right)}{x+1}+\frac{-2x+1}{x+1}\)

=x+\(\frac{-2x-2+3}{x+1}=x+\frac{-2\left(x+1\right)}{x+1}+\frac{3}{x+1}=x-2+\frac{3}{x+1}\in Z\)<=>\(\frac{3}{x+1}\in Z\Leftrightarrow x+1\inƯ\left(3\right)\)

phần này tự làm vì nó dễ

học tốt!