[toán 9] tìm x,y,z

C

chonhoi110

Đặt $\dfrac{x+y\sqrt{2013}}{y+z\sqrt{2013}}=\dfrac{a}{b}$ (với $a,b \in \mathbb{N}* ; (a,b)=1)$

~> $\sqrt{2013}(by-az)=ay-bx$ ~> $ \left\{\begin{matrix}by=az & &\\ ay=bx\end{matrix}\right.$ ~> $\dfrac{x}{y}=\dfrac{y}{z}(=\dfrac{a}{b})$ ~> $y^2=xz$

~> $x^2+y^2+z^2=(x+z)^2-y^2=(x-y+z)(x+y+z)$

Vì $x+y+z >1$ ~> $x^2+y^2+z^2$ là số nguyên tố <~> $\left\{\begin{matrix}x^2+y^2+z^2=x+y+z & &\\ x-y+z=1 \end{matrix}\right.$ ~> $x=y=z=1$
 
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