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[原创] 关于三角形或四面体,我也提一个问题。。。

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发表于 2013-12-1 21:20:09 | 显示全部楼层
由36#可知,若四面面积相等,则有如下结论
QQ截图20131201211943.png

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此表达式处理的很好,但对于一般的关系式,你可以得到吗?见44#回复  发表于 2013-12-8 09:01

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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-6 12:48:24 | 显示全部楼层
四面体为等腰四面体(四面体四个表面积相等)的充要条件是内心与重心重合
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-7 01:26:59 | 显示全部楼层
对于四等分四面体\(ABCD\)体积的点\(P\),设此时\(PA=a,PB=b,PC=c,PD=d\)

我们可以解得四面体各棱长:

\(BC=\sqrt{3a^2+3b^2+3c^2-d^2-y^2-z^2}, AD=\sqrt{5a^2+b^2+c^2+d^2-y^2-z^2}, AC=y,BD=\sqrt{-2a^2+2b^2-2c^2+2d^2+y^2}, AB=z,CD=\sqrt{-2a^2-2b^2+2c^2+2d^2+z^2}\)

并且有关系式:\(AB^2+AC^2+BC^2+AD^2+BD^2+CD^2=4(a^2+b^2+c^2+d^2)\)

四面体\(ABCD\)体积最大时仅当下式成立

\(10a^4+12a^2b^2+7a^2c^2-3a^2d^2-4a^2y^2-7a^2z^2+2b^4+5b^2c^2-b^2d^2-4b^2y^2-3b^2z^2-3c^2z^2+d^2z^2+2y^2z^2+z^4=0\)

\(10a^4+7a^2b^2+12a^2c^2-3a^2d^2-7a^2y^2-4a^2z^2+5b^2c^2-3b^2y^2+2c^4-c^2d^2-3c^2y^2-4c^2z^2+d^2y^2+y^4+2y^2z^2=0\)

\((64a^2+64b^2-32z^2)y^4+(-32z^4+(224a^2+96b^2+96c^2-32d^2)z^2-320a^4-384a^2b^2-224a^2c^2+96a^2d^2-64b^4-160b^2c^2+32b^2d^2)y^2+(64a^2+64c^2)z^4+
(-320a^4-224a^2b^2-384a^2c^2+96a^2d^2-160b^2c^2-64c^4+32c^2d^2)z^2+384a^6+576a^4b^2+576a^4c^2-224a^4d^2+192a^2b^4+480a^2b^2c^2-160a^2b^2d^2+
192a^2c^4-160a^2c^2d^2+32a^2d^4+96b^4c^2+96b^2c^4-32b^2c^2d^2+288v^2=0\)

我们容易得到四等分\(ABCD\)体积的最大值\(v\)满足:

\(-177147v^8+(729s_1^3-26244s_1s_2+157464s_3)v^6+(81s_1^4s_2-2916s_1^3s_3-648s_1^2s_2^2+27216s_1^2s_4+11664s_1s_2s_3+1296s_2^3-46656s_2s_4-34992s_3^2)v^4+
(9s_1^6s_3-360s_1^5s_4-108s_1^4s_2s_3+2880s_1^3s_2s_4+432s_1^2s_2^2s_3-5184s_1^2s_3s_4-5760s_1s_2^2s_4-576s_2^3s_3-13824s_1s_4^2+20736s_2s_3s_4)v^2+s_1^8s_4-
16s_1^6s_2s_4+96s_1^4s_2^2s_4-128s_1^4s_4^2-256s_1^2s_2^3s_4+1024s_1^2s_2s_4^2+256s_2^4s_4-2048s_2^2s_4^2+4096s_4^3=0\)

其中:

\(s_1=a^2+b^2+c^2+d^2\)

\(s_2=a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2\)

\(s_3=a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2\)

\(s_4=a^2b^2c^2d^2\)


毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-8 08:59:22 | 显示全部楼层
不知zeus能否给出四面体及内点的一般关系式,四面体各棱长\(BC=x,AC=y,AB=z,AD=x_1,BD=y_1,CD=z_1\),内点到各顶点的距离为\(PA=a,PB=b,PC=c,PD=d\)? 即类似于41#的对称表达式。
当然我们根据体积关系可以得到

\(f_1=-2a^4x^2+2a^2b^2x^2+2a^2b^2y^2-2a^2b^2z^2+2a^2c^2x^2-2a^2c^2y^2+2a^2c^2z^2-2a^2x^4+2a^2x^2y^2+2a^2x^2z^2-2b^4y^2-
2b^2c^2x^2+2b^2c^2y^2+2b^2c^2z^2+2b^2x^2y^2-2b^2y^4+2b^2y^2z^2-2c^4z^2+2c^2x^2z^2+2c^2y^2z^2-2c^2z^4-2x^2y^2z^2\)

\(f_2=-2a^4y_1^2+2a^2b^2x_1^2+2a^2b^2y_1^2-2a^2b^2z^2-2a^2d^2x_1^2+2a^2d^2y_1^2+2a^2d^2z^2+2a^2x_1^2y_1^2-2a^2y_1^4+2a^2y_1^2z^2-
2b^4x_1^2+2b^2d^2x_1^2-2b^2d^2y_1^2+2b^2d^2z^2-2b^2x_1^4+2b^2x_1^2y_1^2+2b^2x_1^2z^2-2d^4z^2+2d^2x_1^2z^2+2d^2y_1^2z^2-2d^2z^4-2x_1^2y_1^2z^2\)

\(f_3=-2a^4z_1^2+2a^2c^2x_1^2-2a^2c^2y^2+2a^2c^2z_1^2-2a^2d^2x_1^2+2a^2d^2y^2+2a^2d^2z_1^2+2a^2x_1^2z_1^2+2a^2y^2z_1^2-2a^2z_1^4-
2c^4x_1^2+2c^2d^2x_1^2+2c^2d^2y^2-2c^2d^2z_1^2-2c^2x_1^4+2c^2x_1^2y^2+2c^2x_1^2z_1^2-2d^4y^2+2d^2x_1^2y^2-2d^2y^4+2d^2y^2z_1^2-2x_1^2y^2z_1^2\)

\(f_4=-2b^4z_1^2-2b^2c^2x^2+2b^2c^2y_1^2+2b^2c^2z_1^2+2b^2d^2x^2-2b^2d^2y_1^2+2b^2d^2z_1^2+2b^2x^2z_1^2+2b^2y_1^2z_1^2-2b^2z_1^4-2c^4y_1^2+
2c^2d^2x^2+2c^2d^2y_1^2-2c^2d^2z_1^2+2c^2x^2y_1^2-2c^2y_1^4+2c^2y_1^2z_1^2-2d^4x^2-2d^2x^4+2d^2x^2y_1^2+2d^2x^2z_1^2-2x^2y_1^2z_1^2\)

\(f_0=-2x^4x_1^2-2x^2x_1^4+2x^2x_1^2y^2+2x^2x_1^2y_1^2+2x^2x_1^2z^2+2x^2x_1^2z_1^2+2x^2y^2y_1^2-2x^2y^2z^2-2x^2y_1^2z_1^2+2x^2z^2z_1^2+
2x_1^2y^2y_1^2-2x_1^2y^2z_1^2-2x_1^2y_1^2z^2+2x_1^2z^2z_1^2-2y^4y_1^2-2y^2y_1^4+2y^2y_1^2z^2+2y^2y_1^2z_1^2+2y^2z^2z_1^2+2y_1^2z^2z_1^2-2z^4z_1^2-2z^2z_1^4\)

\(\sqrt{f_1}+\sqrt{f_2}+\sqrt{f_3}+\sqrt{f_4}=\sqrt{f_0}\)

此计算已超出了我的计算机处理能力,当然若能确定\(f_0,f_1,f_2,f_3,f_4\)基本简化关系式,将其简化后计算也是可行的。
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-8 11:45:11 | 显示全部楼层
四等分体积:
AB^2+AC^2+AD^2=5*a^2+b^2+c^2+d^2 ;
AB^2+BC^2+BD^2=a^2+5*b^2+c^2+d^2 ;
AC^2+BC^2+CD^2=a^2+b^2+5*c^2+d^2 ;
AD^2+BD^2+CD^2=a^2+b^2+c^2+5*d^2 ;
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-8 11:46:47 | 显示全部楼层
前面得出过正四面体情况解如果存在必然唯一,那么既然面积相等等价于对边相等,必然解也不唯一,有两个自由度。

点评

具体的解答可见43#,我假定了两个自由度 y,z  发表于 2013-12-8 13:06
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-8 13:03:18 来自手机 | 显示全部楼层
而能够有解的充要条件应该还是最大的小于另外三个之和
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-8 20:56:40 来自手机 | 显示全部楼层
充要条件还是比较难找,47楼不对
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2013-12-21 00:37:47 | 显示全部楼层
关于43#的代数方程,其实可以展开为(空间五点的Cayley-Menger行列式定理)得到:(猜想)

\(AB=z,AC=y,BC=x,CD=z_1,BD=y_1,AD=x_1,AP=a,BP=b,CP=c,DP=d\)的一般关系式:

\(-a^4x^4+2a^4x^2y_1^2+2a^4x^2z_1^2-a^4y_1^4+2a^4y_1^2z_1^2-a^4z_1^4-2a^2b^2x^2x_1^2+2a^2b^2x^2y^2-2a^2b^2x^2z_1^2+2a^2b^2x_1^2y_1^2-2a^2b^2x_1^2z_1^2-2a^2b^2y^2y_1^2-
2a^2b^2y^2z_1^2-2a^2b^2y_1^2z_1^2+4a^2b^2z^2z_1^2+2a^2b^2z_1^4-2a^2c^2x^2x_1^2-2a^2c^2x^2y_1^2+2a^2c^2x^2z^2-2a^2c^2x_1^2y_1^2+2a^2c^2x_1^2z_1^2+4a^2c^2y^2y_1^2+
2a^2c^2y_1^4-2a^2c^2y_1^2z^2-2a^2c^2y_1^2z_1^2-2a^2c^2z^2z_1^2+2a^2d^2x^4+4a^2d^2x^2x_1^2-2a^2d^2x^2y^2-2a^2d^2x^2y_1^2-2a^2d^2x^2z^2-2a^2d^2x^2z_1^2-
2a^2d^2y^2y_1^2+2a^2d^2y^2z_1^2+2a^2d^2y_1^2z^2-2a^2d^2z^2z_1^2+2a^2x^4x_1^2-2a^2x^2x_1^2y_1^2-2a^2x^2x_1^2z_1^2-2a^2x^2y^2y_1^2+4a^2x^2y_1^2z_1^2-2a^2x^2z^2z_1^2+
2a^2y^2y_1^4-2a^2y^2y_1^2z_1^2-2a^2y_1^2z^2z_1^2+2a^2z^2z_1^4-b^4x_1^4+2b^4x_1^2y^2+2b^4x_1^2z_1^2-b^4y^4+2b^4y^2z_1^2-b^4z_1^4+4b^2c^2x^2x_1^2+2b^2c^2x_1^4-2b^2c^2x_1^2y^2-
2b^2c^2x_1^2y_1^2-2b^2c^2x_1^2z^2-2b^2c^2x_1^2z_1^2-2b^2c^2y^2y_1^2+2b^2c^2y^2z^2+2b^2c^2y_1^2z_1^2-2b^2c^2z^2z_1^2-2b^2d^2x^2x_1^2-2b^2d^2x^2y^2+2b^2d^2x^2z_1^2-
2b^2d^2x_1^2y^2+2b^2d^2x_1^2z^2+2b^2d^2y^4+4b^2d^2y^2y_1^2-2b^2d^2y^2z^2-2b^2d^2y^2z_1^2-2b^2d^2z^2z_1^2+2b^2x^2x_1^4-2b^2x^2x_1^2y^2-2b^2x^2x_1^2z_1^2-
2b^2x_1^2y^2y_1^2+4b^2x_1^2y^2z_1^2-2b^2x_1^2z^2z_1^2+2b^2y^4y_1^2-2b^2y^2y_1^2z_1^2-2b^2y^2z^2z_1^2+2b^2z^2z_1^4-c^4x_1^4+2c^4x_1^2y_1^2+2c^4x_1^2z^2-c^4y_1^4+
2c^4y_1^2z^2-c^4z^4-2c^2d^2x^2x_1^2+2c^2d^2x^2y_1^2-2c^2d^2x^2z^2+2c^2d^2x_1^2y^2-2c^2d^2x_1^2z^2-2c^2d^2y^2y_1^2-
2c^2d^2y^2z^2-2c^2d^2y_1^2z^2+2c^2d^2z^4+4c^2d^2z^2z_1^2+2c^2x^2x_1^4-2c^2x^2x_1^2y_1^2-2c^2x^2x_1^2z^2-2c^2x_1^2y^2y_1^2+4c^2x_1^2y_1^2z^2-2c^2x_1^2z^2z_1^2+
2c^2y^2y_1^4-2c^2y^2y_1^2z^2-2c^2y_1^2z^2z_1^2+2c^2z^4z_1^2-d^4x^4+2d^4x^2y^2+2d^4x^2z^2-d^4y^4+2d^4y^2z^2-d^4z^4+2d^2x^4x_1^2-2d^2x^2x_1^2y^2-2d^2x^2x_1^2z^2-
2d^2x^2y^2y_1^2+4d^2x^2y^2z^2-2d^2x^2z^2z_1^2+2d^2y^4y_1^2-2d^2y^2y_1^2z^2-2d^2y^2z^2z_1^2+2d^2z^4z_1^2-x^4x_1^4+2x^2x_1^2y^2y_1^2+2x^2x_1^2z^2z_1^2-y^4y_1^4+
2y^2y_1^2z^2z_1^2-z^4z_1^4=0\)            (1)


若四个面面积相等,令\(x_1=x,y_1=y,z_1=z\) 代入(1)可以得到

\(-a^4x^4+2a^4x^2y^2+2a^4x^2z^2-a^4y^4+2a^4y^2z^2-a^4z^4-2a^2b^2x^4+4a^2b^2x^2y^2-4a^2b^2x^2z^2-2a^2b^2y^4-4a^2b^2y^2z^2+6a^2b^2z^4-2a^2c^2x^4-4a^2c^2x^2y^2+
4a^2c^2x^2z^2+6a^2c^2y^4-4a^2c^2y^2z^2-2a^2c^2z^4+6a^2d^2x^4-4a^2d^2x^2y^2-4a^2d^2x^2z^2-2a^2d^2y^4+4a^2d^2y^2z^2-2a^2d^2z^4+2a^2x^6-2a^2x^4y^2-2a^2x^4z^2-
2a^2x^2y^4+4a^2x^2y^2z^2-2a^2x^2z^4+2a^2y^6-2a^2y^4z^2-2a^2y^2z^4+2a^2z^6-b^4x^4+2b^4x^2y^2+2b^4x^2z^2-b^4y^4+2b^4y^2z^2-b^4z^4+6b^2c^2x^4-4b^2c^2x^2y^2-
4b^2c^2x^2z^2-2b^2c^2y^4+4b^2c^2y^2z^2-2b^2c^2z^4-2b^2d^2x^4-4b^2d^2x^2y^2+4b^2d^2x^2z^2+6b^2d^2y^4-4b^2d^2y^2z^2-2b^2d^2z^4+2b^2x^6-2b^2x^4y^2-2b^2x^4z^2-
2b^2x^2y^4+4b^2x^2y^2z^2-2b^2x^2z^4+2b^2y^6-2b^2y^4z^2-2b^2y^2z^4+2b^2z^6-c^4x^4+2c^4x^2y^2+2c^4x^2z^2-c^4y^4+2c^4y^2z^2-c^4z^4-2c^2d^2x^4+4c^2d^2x^2y^2-
4c^2d^2x^2z^2-2c^2d^2y^4-4c^2d^2y^2z^2+6c^2d^2z^4+2c^2x^6-2c^2x^4y^2-2c^2x^4z^2-2c^2x^2y^4+4c^2x^2y^2z^2-2c^2x^2z^4+2c^2y^6-2c^2y^4z^2-2c^2y^2z^4+2c^2z^6-
d^4x^4+2d^4x^2y^2+2d^4x^2z^2-d^4y^4+2d^4y^2z^2-d^4z^4+2d^2x^6-2d^2x^4y^2-2d^2x^4z^2-2d^2x^2y^4+4d^2x^2y^2z^2-2d^2x^2z^4+2d^2y^6-2d^2y^4z^2-2d^2y^2z^4+2d^2z^6-
x^8+2x^4y^4+2x^4z^4-y^8+2y^4z^4-z^8=0\)  (2)

很奇怪:41#得到的答案:

\(x^8+y^8+z^8-2x^4y^4-2y^4z^4-2z^4x^4+(a^4+b^4+c^4+d^4)(x^4-2x^2y^2-2x^2z^2+y^4-2y^2z^2+z^4)+(2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2)
(x^4+2x^2y^2+2x^2z^2+y^4+2y^2z^2+z^4)+(2(a^2+b^2+c^2+d^2))(-x^6+x^4y^2+x^4z^2+x^2y^4-2x^2y^2z^2+x^2z^4-y^6+y^4z^2+y^2z^4-z^6)-(8(a^2b^2+c^2d^2))(x^4+y^2z^2)-
(8(a^2c^2+b^2d^2))(x^2z^2+y^4)-(8(a^2d^2+b^2c^2))(x^2y^2+z^4)=0\)   (3)  与(2)并不相等,

当\(x=y=z=1,a=b=c=d=\frac{\sqrt{6}}{4}\)时,(2)和(3)均成立,是否(3)计算有误?

若为垂心四面体,设\(x^2+x_1^2=y^2+y_1^2=z^2+z_1^2=t^2\)则

\((-x^4+2x^2y^2+2z^2x^2-y^4+2y^2z^2-z^4)t^4+(4a^4x^2-4a^2b^2x^2-4a^2b^2y^2+4a^2b^2z^2-4a^2c^2x^2+4a^2c^2y^2-4a^2c^2z^2+6a^2x^4-4a^2x^2y^2-4a^2x^2z^2-2a^2y^4+4a^2y^2z^2-2a^2z^4+4b^4y^2+4b^2c^2x^2-
4b^2c^2y^2-4b^2c^2z^2-2b^2x^4-4b^2x^2y^2+4b^2x^2z^2+6b^2y^4-4b^2y^2z^2-2b^2z^4+4c^4z^2-2c^2x^4+4c^2x^2y^2-4c^2x^2z^2-2c^2y^4-4c^2y^2z^2+6c^2z^4+2d^2x^4-
4d^2x^2y^2-4d^2x^2z^2+2d^2y^4-4d^2y^2z^2+2d^2z^4+2x^6-2x^4y^2-2x^4z^2-2x^2y^4-2x^2z^4+2y^6-2y^4z^2-2y^2z^4+2z^6)t^2+4a^2c^2x^2z^2-4b^2d^2x^2z^2+4b^2x^2y^2z^2-4a^2d^2y^2z^2+4a^2x^2y^2z^2-2a^2d^2x^4-2a^2b^2z^4-4c^2d^2x^2y^2+4c^2x^2y^2z^2-c^4z^4-d^4x^4-d^4y^4-d^4z^4-
2a^2x^6+2a^2y^6+2a^2z^6-b^4x^4-b^4z^4+2b^2x^6-2b^2y^6+2b^2z^6-c^4x^4-c^4y^4+2c^2x^6+2c^2y^6-2c^2z^6-2d^2x^6-2d^2y^6-2d^2z^6+2a^2c^2x^4+2x^4y^4+2z^4x^4+2y^4z^4-
a^4x^4-a^4y^4-a^4z^4+2a^2d^2y^4+2a^2d^2z^4-2a^2x^4y^2-2a^2x^4z^2+2a^2x^2y^4+2a^2x^2z^4-2a^2y^4z^2-2a^2y^2z^4-2b^4x^2y^2+2b^4x^2z^2-2b^4y^2z^2+2a^2b^2x^4-
b^4y^4+2b^2c^2z^4+2b^2d^2x^4+2b^2d^2z^4+2b^2x^4y^2-2b^2x^4z^2-2b^2x^2y^4-2b^2x^2z^4-2b^2y^4z^2+2b^2y^2z^4+2c^4x^2y^2-2c^4x^2z^2-2c^4y^2z^2+2c^2d^2x^4+
2c^2d^2y^4-2c^2x^4y^2+2c^2x^4z^2-2c^2x^2y^4-2c^2x^2z^4+2c^2y^4z^2-2c^2y^2z^4+2d^2x^4y^2+2d^2x^4z^2-2a^2c^2y^4+2a^2b^2y^4-2a^4x^2z^2+2a^4y^2z^2-2a^4x^2y^2+
2a^2c^2z^4+2d^2x^2y^4+2d^2x^2z^4+2d^2y^4z^2+2d^2y^2z^4-x^8-y^8-z^8+4a^2b^2x^2y^2+4d^2x^2y^2z^2+4b^2c^2y^2z^2-2b^2c^2x^4+2b^2c^2y^4+2d^4x^2z^2+2d^4y^2z^2-
2b^2d^2y^4-2c^2d^2z^4+2d^4x^2y^2=0\)


若为正四面体,即\(x=y=z=x_1=y_1=z_1=t\),则

\(3t^4+(-2a^2-2b^2-2c^2-2d^2)t^2+3a^4-2a^2b^2-2a^2c^2-2a^2d^2+3b^4-2b^2c^2-2b^2d^2+3c^4-2c^2d^2+3d^4=0\)

若\(x_1\*x=y_1\*y=z_1\*z=t\),则(注:此特例即http://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=5190&extra=page%3D2&page=6

mathe推导的58#的结论)

\((-a^4x^4y^4+2a^4x^4y^2z^2-a^4x^4z^4+2a^2b^2x^4y^4-2a^2b^2x^4y^2z^2-2a^2b^2x^2y^4z^2+2a^2b^2x^2y^2z^4-2a^2c^2x^4y^2z^2+2a^2c^2x^4z^4+2a^2c^2x^2y^4z^2-
2a^2c^2x^2y^2z^4+4a^2x^6y^2z^2-2a^2x^4y^4z^2-2a^2x^4y^2z^4-b^4x^4y^4+2b^4x^2y^4z^2-b^4y^4z^4+2b^2c^2x^4y^2z^2-2b^2c^2x^2y^4z^2-2b^2c^2x^2y^2z^4+2b^2c^2y^4z^4-
2b^2x^4y^4z^2+4b^2x^2y^6z^2-2b^2x^2y^4z^4-c^4x^4z^4+2c^4x^2y^2z^4-c^4y^4z^4-2c^2x^4y^2z^4-2c^2x^2y^4z^4+4c^2x^2y^2z^6+3x^4y^4z^4)t^4+(2a^4x^6y^4z^2+2a^4x^6y^2z^4-2a^2b^2x^6y^4z^2-2a^2b^2x^4y^6z^2-2a^2c^2x^6y^2z^4-2a^2c^2x^4y^2z^6-2a^2d^2x^6y^4z^2-2a^2d^2x^6y^2z^4+2a^2d^2x^4y^6z^2+
2a^2d^2x^4y^2z^6-2a^2x^6y^4z^4+2b^4x^4y^6z^2+2b^4x^2y^6z^4-2b^2c^2x^2y^6z^4-2b^2c^2x^2y^4z^6+2b^2d^2x^6y^4z^2-2b^2d^2x^4y^6z^2-2b^2d^2x^2y^6z^4+
2b^2d^2x^2y^4z^6-2b^2x^4y^6z^4+2c^4x^4y^2z^6+2c^4x^2y^4z^6+2c^2d^2x^6y^2z^4-2c^2d^2x^4y^2z^6+2c^2d^2x^2y^6z^4-2c^2d^2x^2y^4z^6-2c^2x^4y^4z^6-
2d^2x^6y^4z^4-2d^2x^4y^6z^4-2d^2x^4y^4z^6t^2-a^4x^8y^4z^4+2a^2b^2x^6y^6z^4+2a^2c^2x^6z^6y^4+2a^2d^2x^8y^4z^4-2a^2d^2x^6y^6z^4-2a^2d^2x^6z^6y^4-
b^4y^8z^4x^4+2b^2c^2y^6z^6x^4-2b^2d^2x^6y^6z^4+2b^2d^2y^8z^4x^4-2b^2d^2y^6z^6x^4-c^4z^8y^4x^4-2c^2d^2x^6z^6y^4-2c^2d^2y^6z^6x^4+2c^2d^2z^8y^4x^4-
d^4x^8y^4z^4+2d^4x^6y^6z^4+2d^4x^6z^6y^4-d^4y^8z^4x^4+2d^4y^6z^6x^4-d^4z^8y^4x^4+4d^2x^6y^6z^6=0\)
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