x^2+y^2+z^2-2x+4y-6y+14=0,求x+y+z 求助,谢谢
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 03:06:08
x^2+y^2+z^2-2x+4y-6y+14=0,求x+y+z 求助,谢谢
x^2+y^2+z^2-2x+4y-6y+14=0,求x+y+z 求助,谢谢
x^2+y^2+z^2-2x+4y-6y+14=0,求x+y+z 求助,谢谢
14=1+4+9
所以(x²-2x+1)+(y²+4y+4)+(z²-6z+9)=0
(x-1)²+(y+2)²+(z-3)²=0
平方相加为0则都等于0
所以x-1=0,y+2=0,z-3=0
x=1,y=-2,z=3
x+y+z=2
因为:x^2-2x+1+y^2+4y+4+z^2-6z+9=0
即:(x-1)^2+(y+2)^2+(z-3)^3=0
所以:x=1;Y=-2;z=3;
所以:x+y+z=2
你打错了一个,是-6Z吧???不然就是变式成:(x^2-2x+1)+(y^2+4y+4)+(z^2-6Z+6)=0
(x+1)^2,+(y-2)^2+(z+3)^2=0,因为平方大于等于0,所以x=-1,y=2,z=-3..
所以x+y+z=-2。
化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z)
计算题(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z).
1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z)
{(z+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y)
(x-y)^2+4z(x-y)+4z^2
求化简(x+3y+2z)(x-3y+6z)(x+3y+4z-2z)(x-3y+4z+2z)
2(x+y-z)-3(x-y+z)-5(x+y-z)+6(x-y+z) 计算
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答
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)
若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-z)=
(x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z
x/6=y/4=z/3,则(x+3y)/(3y-2z)
用行列式的性质证明:y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证?