如何化简 1/(a+b√2+c√3+d√6)

zeroieme

hujunhua 发表于 2021-11-18 10:27
按4#的链接，分母有理化为\[
(a+b\sqrt2+c\sqrt3+d\sqrt6)\*(a+b\sqrt2+c\sqrt3-d\sqrt6)\*(a+b\sqrt2-c\s ...

题由就是  Q(√2,√3) = { a+b√2+c√3+d√6｜a,b,c,d∈Q } 构成一个数域
故这里只有两素数平方根。

葡萄糖

hujunhua 发表于 2021-11-18 10:27
按4#的链接，分母有理化为
\begin{gather*}
(a+b\sqrt2+c\sqrt3+d\sqrt6)\cdot(a+b\sqrt2+c\sqrt3-d\sqrt6)\cdot(a+b\sqrt2-c\sqrt3+d\sqrt6)\cdot(a-b\sqrt2+c\sqrt3+d\sqrt6)\\
\cdot(a-b\sqrt2-c\sqrt3-d\sqrt6)\cdot(a-b\sqrt2-c\sqrt3+d\sqrt6)\cdot(a-b\sqrt2+c\sqrt3-d\sqrt6)\cdot(a+b\sqrt2-c\sqrt3-d\sqrt6)\\
=\left(a^4-4a^2b^2+4b^4-6a^2c^2-12b^2c^2+9c^4-12a^2d^2-24b^2d^2-36c^2d^2+36d^4\right)^2-(48abcd)^2
\end{gather*}
...

\begin{align*}
&&p&=\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\\  
&&q&=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\\  
&\Rightarrow&\left[\left(p^2-\sigma_1\right)^2-4\sigma_2\right]^2&=64p^2\sigma_3\\
&\Rightarrow&\left[\left(p^2-x_1-x_2-x_3\right)^2-4\left(x_1x_2+x_2x_3+x_3x_1\right)\right]^2&=64p^2x_1x_2x_3\\
&\Rightarrow&\left\{\left[(q-a)^2-2b^2-3c^2-6d^2\right]^2-4\left(6b^2c^2+18c^2d^2+12b^2d^2)\right]\right\}^2&=64(q-a)^2\cdot36b^2c^2d^2\\
\end{align*}

因此

\begin{gather*}
(a+b\sqrt2+c\sqrt3+d\sqrt6)\cdot(a+b\sqrt2+c\sqrt3-d\sqrt6)\cdot(a+b\sqrt2-c\sqrt3+d\sqrt6)\cdot(a-b\sqrt2+c\sqrt3+d\sqrt6)\\
\cdot(a-b\sqrt2-c\sqrt3-d\sqrt6)\cdot(a-b\sqrt2-c\sqrt3+d\sqrt6)\cdot(a-b\sqrt2+c\sqrt3-d\sqrt6)\cdot(a+b\sqrt2-c\sqrt3-d\sqrt6)\\
=\left[(a^2-2b^2-3c^2-6d^2)^2-24(b^2c^2+2b^2d^2+3c^2d^2)\right]^2-(48abcd)^2
\end{gather*}