新符号学 (第20/28页)

ac{c}{d}}-x=d_{2}}{\dispystyle{\frac{c}{d}}-x=d_{2}}

    如果有正整数m,k满足：{\dispystyle{\frac{kd}{mb}}={\frac{d_{1}}{d_{2}}}}{\dispystyle{\frac{kd}{mb}}={\frac{d_{1}}{d_{2}}}}

    那麽就有：{\dispystylex={\frac{ma kc}{mb kd}}}{\dispystylex={\frac{ma kc}{mb kd}}}

    证明如下：由条件可得

    {\dispystyle{\begin{aligned}bd_{1}&=bx-a\\dd_{2}&=c-dx\end{aligned}}}{\dispystyle{\begin{aligned}bd_{1}&=bx-a\\dd_{2}&=c-dx\end{aligned}}}

    而根据{\dispystyle{\frac{kd}{mb}}={\frac{d_{1}}{d_{2}}}}{\dispystyle{\frac{kd}{mb}}={\frac{d_{1}}{d_{2}}}}又有

    {\dispystylembd_{1}=kdd_{2}}

    代入上面的两个关系式可得：

    {\dispystylembx-a=kc-dx}

    2<