搜搜考试网若z=x^2 3xy
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1=(x-z-2ab)/xy2=(a²-2ab+b²)/a-b=(a-b)²/a-b=a-
题目出错了吧应该是x^2-4x+y^2+6y+√(z-3)+13=0即（x-2)^2+(y+3)^2+√(z-3)=0(xy)^z=(2*(-3))^3=-216啊你也去想想吧
设(y+z)/x=(z+x)/y=(y+x)/z=k则y+z=kx,z+x=ky,y+x=kz三式相加2(x+y+z)=k(x+y+z)故当x+y+z=0时,k=-1,但xy-z不等于0,可知x+y+
x²+y²+xy=x²+y²-2xycos120度同理y²+z²+yz=y²+z²-2yzcoa120度x²+
若3/x=2/y=5/z,则x/3=y/2=z/5,设x/3=y/2=z/5=K,则x=3K,y=2K,z=5K将其代入上式,得6k2+10k2+15k231k2-------------------
y=6-x所以z²=6x-x²+9(x-3)²+z²=0所以x-3=0,且z=0所以z=0
∵x+y+z=5∴x=5-y-z∵xy+yz+xz=3∴y^2+(z-5)y+(z^2-5z+3)=0又∵y,z是实数,∴△=(z-5)^2-4(z^2-5z+3)=(z+1)(-3z+13)≥0∴-
(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&
求z的值吗y=6-xz²=xy-9=x(6-x)-9=-x²+6x-9z²=-(x-3)²z²+(x-3)²=0所以z=0,x-3=0解得z
2x²+2xy+y²-4x+z-2√z-3+2=0对其化简:(x²+2xy+y²)+(x²-4x+4)+(z-2√z-3-2)=0(x+y)²
可分解为（x+y)²=0.（x-2)²=0)²=0解得x=-yx=2√z-3=1解得x=2y=-2z=4xy=-4yz=-8xz=8
xy+yz+zx=93中的y,z全用x代替可以得到2x^2/3+10x^2/9+5x^2/3=93∴x^2=27同理y^2=12z^2=75∴9x*x+12y*y+2z*z=9*27+12*12+2*
2X-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z代入(1)y=3z结果代入(x^2+3xY)÷(y^2+z^2)=(25z^2+3*5z*3z)÷(9z^2+z
xy=x+y,yz=2y+2z,xz=3x+3z1/x+1/y=1（1）1/y+1/z=1/2（2）1/x+1/z=1/3（3）（1）-（2）1/x-1/z=1/2（4）（3）+（4）2/x=5/6x
(x+y+z)²=6²x²+y²+z²+2(xy+yz+zx)=36x²+y²+z²=36-22=14x³+y
xy+xz+yz=76,x/3=y=z/4所以,19x^2/9=76,x^2=362x*x+12y*y+9z*z=58x^2/3=58*12=696
因为x+y=8,所以x=8-y所以y（8-y）+z平方+16=0又因为z平方=-16-xy,所以y（8-y）-16-xy=0即8y-y平方-16-xy=0,所以（y-4）平方+xy=0所以z平方=（y
｜x-3｜+｜y+z｜+｜2z+1｜=0则｜x-3｜=0x=3｜y+z｜=0y=-z=1/2｜2z+1｜=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4