[Toán 8]Tìm x,y nguyên

H

harrypham

xyx+y=23    3xy=2x+2y    2y=3xy2x    2y=x(3y2)    x=2y3y2\dfrac{xy}{x+y}= \dfrac{2}{3} \implies 3xy=2x+2y \implies 2y=3xy-2x \implies 2y=x(3y-2) \implies x= \dfrac{2y}{3y-2}
Nhận thấy xZ    3xZ    6y3y2Zx \in \mathbb{Z} \iff 3x \in \mathbb{Z} \iff \dfrac{6y}{3y-2} \in \mathbb{Z}.
Phân tích 3x=2(3y2)+43y2=2+43y2Z    3y2{±1,±4,±2}3x= \dfrac{2(3y-2)+4}{3y-2}=2+ \dfrac{4}{3y-2} \in \mathbb{Z} \iff 3y-2 \in \{ \pm 1, \pm 4, \pm 2 \}
    3y{1,3,6,2,4,0}    y{1,2}\iff 3y \in \{ 1,3,6,-2,4,0 \} \iff y \in \{1,2 \}
+ Với y=1    x=2y=1 \implies x=2.
+ Với y=2    x=1y=2 \implies x=1.
Thử lại coi kết quả có đúng không.
 
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