求解方程组：c=b^2／a ，b^2-y^2=c^2-z^2 ，a／y=(a+b-x)／x ，a／z=(a+b-x)／(x+b+c)

数学小白新

求解方程组

\(\begin{cases}
c=\frac{b^2}{a}\\
b^2-y^2=c^2-z^2\\
\frac{a}{y}=\frac{a+b-x}{x}\\
\frac{a}{z}=\frac{a+b-x}{x+b+c}\\
\end{cases}\\\)

Treenewbee

齐次方程组，消元 可得到一组解：

\[\{a,b,c,x,y,z\}=\left\{-\frac{1}{4} \left(\sqrt{5}+\sqrt{2 \left(\sqrt{5}+1\right)}-1\right) t,\frac{1}{2} \left(\sqrt{5}-1\right) t,-\frac{1}{2} \left(\sqrt{5}+\sqrt{2 \left(\sqrt{5}-1\right)}-3\right) t,-\frac{1}{2} \left(\sqrt{2 \left(\sqrt{5}+1\right)}+2\right) t,\frac{1}{2} \left(\sqrt{5}+2 \sqrt{\sqrt{5}-2}-1\right) t,t\right\}\]

Treenewbee

\[p\ne q,\left\{\frac{\text{pq}}{q-p},0,0,p,q,q\right\}\]
\[p\ne 0,\{-q,q,-q,p,q,q\}\]
\[p\ne q,\left\{\frac{p^2}{q},p,q,\frac{p (p+q)}{q-p},p,q\right\}\]

luyuanhong

楼上 Treenewbee 的解答已收藏。