Question

Tìm các số nguyên x, y thỏa mãn x³+3x=x²y+2y+5

Answer 1

\(x^3+3x=x^2y+2y+5\\ \Leftrightarrow x^3+3x-x^2y-2y-5=0\\ \Leftrightarrow\left(x^3+2x\right)-y\left(x^2+2\right)+x-5=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-y\right)=5-x\)

ĐỀ có sao ko!!

Answer 2

x3+3x=x2y+2y+5

⇔x3+3x−x2y−2y−5=0

⇔(x3+2x)−y(x2+2)+x−5=0

⇔(x2+2)(x−y)=5−x