Question

Tìm nghiệm nguyên của phương trình:
(x² + 1)y = x

Answer 1

\(\left(x^2+1\right)y=x\Rightarrow y=\dfrac{x}{x^2+1}\)

Do \(\dfrac{x}{x^2+1}=\dfrac{\dfrac{1}{2}\left(x^2+2x+1\right)-\dfrac{1}{2}\left(x^2+1\right)}{x^2+1}=\dfrac{\left(x+1\right)^2}{2\left(x^2+1\right)}-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

Và \(\dfrac{x}{x^2+1}=\dfrac{\dfrac{1}{2}\left(x^2+1\right)-\dfrac{1}{2}\left(x^2-2x+1\right)}{x^2+1}=\dfrac{1}{2}-\dfrac{\left(x-1\right)^2}{2\left(x^2+1\right)}\le\dfrac{1}{2}\)

\(\Rightarrow-\dfrac{1}{2}\le y\le\dfrac{1}{2}\)

Mà \(y\in Z\Rightarrow y=0\)

\(\Rightarrow x=0\)