且恰好通过如图所示的隧道（上部为半圆形）卡车有多高？

设x+y+z=0且x2/(b-c

设x+y+z=0且x2/(b-c)+y2/(c-a)+z2/(a-b)=0求(a2x+b2y+c2z)/(bcx+acy+abz) 已知x+y+z=0 所以，z=-(x+y) 则，z^2=(x+y)^2 又，x^2/(b-c)+y^2/(c-a)+z^2/(a-b)=0 ===> (c-a)(a-b)x^2+(b-c)(a-b)y^2+(b-c)(c-a)(x+y)^2=0 ===> -(c-a)*[(c-a)+(b-c)]x^2-(b-c)*[(c-a)+(b-c)]y^2+(b-c)(c-a)[(x^2+y^2)+2xy]=0 ===> -(c-a)^2x^2-(b-c)(c-a)x^...全部

  设x+y+z=0且x2/(b-c)+y2/(c-a)+z2/(a-b)=0求(a2x+b2y+c2z)/(bcx+acy+abz) 已知x+y+z=0 所以，z=-(x+y) 则，z^2=(x+y)^2 又，x^2/(b-c)+y^2/(c-a)+z^2/(a-b)=0 ===> (c-a)(a-b)x^2+(b-c)(a-b)y^2+(b-c)(c-a)(x+y)^2=0 ===> -(c-a)*[(c-a)+(b-c)]x^2-(b-c)*[(c-a)+(b-c)]y^2+(b-c)(c-a)[(x^2+y^2)+2xy]=0 ===> -(c-a)^2x^2-(b-c)(c-a)x^2-(b-c)^2y^2-(b-c)(c-a)y^2+(b-c)(c-a)(x^2+y^2)+2(b-c)(c-a)xy=0 ===> -(c-a)^2x^2-(b-c)^2y^2+2(b-c)(c-a)xy=0 ===> (c-a)^2x^2-2(b-c)(c-a)xy+(b-c)^2y^2=0 ===> [(c-a)x-(b-c)y]^2=0 ===> (c-a)x=(b-c)y ===> x=[(b-c)/(c-a)]y 代入到x+y+z=0中得到：[(b-c)/(c-a)]y+y+z=0 ===> [(b-c)+(c-a)]y/(c-a)+z=0 ===> [(b-a)/(c-a)]y+z=0 ===> z=[(a-b)/(c-a)]y 再将x=[(b-c)/(c-a)]y，y=y，z=[(a-b)/(c-a)]y代入表达式 (a^2x+b^2y+c^2z)/(bcx+acy+abz)中就有： 原式=1。
  收起