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Question

Tìm x,y nguyên, biết x²-xy=6x-5y-8

anbe · Accepted Answer

Pt$\Leftrightarrow$$y\left(x-5\right)=x^2-6x+8$

$\Leftrightarrow y=\dfrac{x^2-6x+8}{x-5}=\dfrac{x\left(x-5\right)-\left(x-5\right)+3}{x-5}=x-1+\dfrac{3}{x-5}$

Để y nguyên $3⋮x-5$ $\Leftrightarrow x-5\inƯ\left(3\right)$

$\Rightarrow x\in\left\{2;3;4;6;7;8\right\}$

Vậy Pt có cặp nghiệm (x,y)={(2;-2),{4;0),(6;8),(8,8)}

Câu trả lời:

Pt$\Leftrightarrow$$y\left(x-5\right)=x^2-6x+8$

$\Leftrightarrow y=\dfrac{x^2-6x+8}{x-5}=\dfrac{x\left(x-5\right)-\left(x-5\right)+3}{x-5}=x-1+\dfrac{3}{x-5}$

Để y nguyên $3⋮x-5$ $\Leftrightarrow x-5\inƯ\left(3\right)$

$\Rightarrow x\in\left\{2;3;4;6;7;8\right\}$

Vậy Pt có cặp nghiệm (x,y)={(2;-2),{4;0),(6;8),(8,8)}