作业帮(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求：xy+yz+zx/x^2+y^2+z^2
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/26 16:22:28
xRJ@~ SIBG)HnNE[""U
J/#I{+8M Cog8"QBTdc\]&ާ%L) F|^,z=H«< JGli#tpĪ |_;Gz?iPUVd{iq'Q ?yȯI$lAdw?@6
L)h5
=q-,&ϯ٫zƊar 5 YNQ?oϻr *$&Dgd7q`Rn)!
-7 unwIU`3q36%u.sy> kxcI觏 k/|r1
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求：xy+yz+zx/x^2+y^2+z^2
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求：xy+yz+zx/x^2+y^2+z^2
(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 xy+yz+zx/x^2+y^2+z^2(x-3y+z)^2+/ 5x-4y+z/=0 且xyz≠0 求：xy+yz+zx/x^2+y^2+z^2
平方和绝对值都大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个式子都等于0
所以x-3y+z=0 (1)
5x-4y+z=0 (2)
(1)-(1)
4x-y=0
y=4x
(2)-(1)*5
5x-4y+z-5x+15y-5z=0
11y-4z=0
z=11y/4,y=4x
所以z=11x
代入xy+yz+zx/x^2+y^2+z^2
=(4x^2+44x^2+11x^2)/(x^2+16x^2+121x^2)
=59x^2/138x^2
=59/138
(x-3y+z)^2和/ 5x-4y+z/都非负，且相加为0
所以x-3y+z=5x-4y+z=0
于是y=4x ,z=11x
所以xy+yz+zx/x^2+y^2+z^2=
（4+44+11）x^2/（1+16+121）x^2=59/138
(4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=?
设|x|-|y|+|z|=|x+y+z|,且|x+y|=4,|y+z|=5,|y|=3,求(x-y-z)^2的值
(4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊
2(x+y-z)-3(x-y+z)-5(x+y-z)+6(x-y+z) 计算
化简-(-3x-y+2z)-{-x+【4x-(x-y-z)-3x+2z}
x/2=y/3=z/5 x+3y-z/x-3y+z
x+2y=3x+2z=4y+z 求x:y:z
x+y/2=z+x/3=y+z/4 x+y+z=27
数学--整式运算4(x-y+z)-2(x+y-z)-3(-x-y-z)
x+y=5 2x+3y-z=9 x-y+z=4
1.（x+y-z)+(x-y+z)-(x-y-z) 2.2x-(3x-2y+3)-(5y-2)
已知：4x-5y+2z=0 x+4y-3z=0则：(x*x+y*y+z*z)/(x*y+y*z+z*x)=____.
求化简(x+3y+2z)(x-3y+6z)(x+3y+4z-2z)(x-3y+4z+2z)
5x+y-2z 4 x+2y-3z 2x-3y+5z三元一次方程组
3x-2y+z=4 x+y-z=5 4x-4y+3z=27
｛ 5x-3y+z=25x+2y-4z=3-5x+y-z=2
x+y+z=4 2x-y+z=3 3x-2y-3z=-5
x:y:z=3:4:5 2x-3y+2z=8 x-y-2z=?