1/2(x+y+z)^2+1/2[(x-y)^2-2z(x-y)+z^2]-z(x+y)
=1/2(x+y)^2+z(x+y)+1/2*z^2+1/2(x-y)^2-z(x-y)+1/2*z^2-z(x+y)
=1/2(x+y)^2+1/2(x-y)^2+z^2-z(x-y)
=1/2(x^2-2xy+y^2+4xy)+1/2(x-y)^2+z^2-z(x-y)
=1/2(x-y)^2+2xy+1/2(x-y)^2+z^2-z(x-y)
=(x-y)^2+2xy-z(x-y)+z^2
代入,x-y=6,xy=21
原式=6^2+2*21-6z+z^2
=36+42-6z+z^2
=78-6z+z^2
不好意思,只能消到此步了。如果原题确实只有这个条件,
那我能力有限。
呵呵z我消不掉了。
如果有其他条件,可以求出数值解的
不好意思~呵呵
1/2(x+y+z)^2+1/2(x-y-z)(x-y+z)-z(x+y)
=[(x+y)^2+2z(x+y)+z^2]/2+[(x-y)^2-z^2]/2-[2z(x+y)]/2
=[(x+y)^2+2z(x+y)+z^2+(x-y)^2-z^2-2z(x+y)]/2
=[(x+y)^2+(x-y)^2]/2
=[x^2+y^2+2xy+(x-y)^2]/2
=[x^2+y^2-2xy+4xy+(x-y)^2]/2
=[(x-y)^2+4xy+(x-y)^2]/2
=[36+84+36]/2
=36+42
=78
cxlovewh:抄别人答案也要分辩对错啊
1/2(x+y+z)^2+1/2[(x-y)^2-2z(x-y)+z^2]-z(x+y)
=1/2(x+y)^2+z(x+y)+1/2*z^2+1/2(x-y)^2-z(x-y)+1/2*z^2-z(x+y)
=1/2(x+y)^2+1/2(x-y)^2+z^2-z(x-y)
=1/2(x^2-2xy+y^2+4xy)+1/2(x-y)^2+z^2-z(x-y)
=1/2(x-y)^2+2xy+1/2(x-y)^2+z^2-z(x-y)
=(x-y)^2+2xy-z(x-y)+z^2
代入,x-y=6,xy=21
原式=6^2+2*21-6z+z^2
=36+42-6z+z^2
=78-6z+z^2
不好意思,只能消到此步了。如果原题确实只有这个条件,
那我能力有限。
呵呵z我消不掉了。
如果有其他条件,可以求出数值解的
不好意思~呵呵
回答者:Sylvia_yaya - 高级魔法师 六级 8-1 02:10
1/2(x+y+z)^2+1/2[(x-y)^2-2z(x-y)+z^2]-z(x+y)
=1/2(x+y)^2+z(x+y)+1/2*z^2+1/2(x-y)^2-z(x-y)+1/2*z^2-z(x+y)
=1/2(x+y)^2+1/2(x-y)^2+z^2-z(x-y)
=1/2(x^2-2xy+y^2+4xy)+1/2(x-y)^2+z^2-z(x-y)
=1/2(x-y)^2+2xy+1/2(x-y)^2+z^2-z(x-y)
=(x-y)^2+2xy-z(x-y)+z^2
代入x-y=6,xy=21
原式=6^2+2*21-6z+z^2
=36+42-6z+z^2
=78-6z+z^2
代入z求值
回答者:cxlovewh - 经理 四级 8-1 02:14
1/2(x+y+z)^2+1/2(x-y-z)(x-y+z)-z(x+y)
=[(x+y)^2+2z(x+y)+z^2]/2+[(x-y)^2-z^2]/2-[2z(x+y)]/2
=[(x+y)^2+2z(x+y)+z^2+(x-y)^2-z^2-2z(x+y)]/2
=[(x+y)^2+(x-y)^2]/2
=[x^2+y^2+2xy+(x-y)^2]/2
=[x^2+y^2-2xy+4xy+(x-y)^2]/2
=[(x-y)^2+4xy+(x-y)^2]/2
=[36+84+36]/2
=36+42
=78
cxlovewh:抄别人答案也要分辩对错啊
1/2(x+y+z)^2+1/2[(x-y)^2-2z(x-y)+z^2]-z(x+y)
=1/2(x+y)^2+z(x+y)+1/2*z^2+1/2(x-y)^2-z(x-y)+1/2*z^2-z(x+y)
=1/2(x+y)^2+1/2(x-y)^2+z^2-z(x-y)
=1/2(x^2-2xy+y^2+4xy)+1/2(x-y)^2+z^2-z(x-y)
=1/2(x-y)^2+2xy+1/2(x-y)^2+z^2-z(x-y)
=(x-y)^2+2xy-z(x-y)+z^2
代入x-y=6,xy=21
原式=6^2+2*21-6z+z^2
=36+42-6z+z^2
=78-6z+z^2
代入z求值
好奇怪的题