代数式恒等变形

Tesla35

(2007上海市初中数学竞赛)已知实数$a,b,c$,且$b\neq0$.若实数$x_1,x_2,y_1,y_2$满足$x_1^2+ax_2^2=b,x_2y_1-x_1y_2=a,x_1y_1+ax_2y_2=c$,则$y_1^2+ay_2^2$的值为

这题考查什么知识?

色k

第一感觉：拉格朗日恒等式

hbghlyj

又见Pell's Equation
A very useful way to combine solutions of the equation into new solutions is furnished by Brahmagupta's identity
\[
(x^2-ny^2)(a^2-nb^2) = (xa+nyb)^2 - n(xb+ya)^2.
\]

hbghlyj

normed field $\Bbb R[\sqrt{-n}]=\{x+y\sqrt{-n}|x,y\in\Bbb R\}$
The case $n=1$ is $\Bbb C$
\begin{align*}\|x+y\sqrt{-n}\|\cdot\|a+b\sqrt{-n}\|&=(x^2+ny^2)(a^2+nb^2)
\\\|(x+y\sqrt{-n})\cdot(a+b\sqrt{-n})\|&=\|xa-nyb+(xb+ya)\sqrt{-n}\|=(xa-nyb)^2+n(xb+ya)^2\end{align*}

isee

色k 发表于 2022-11-11 17:24
第一感觉：拉格朗日恒等式

还有点评公式，强大