z=x平方-xy y方-2x y 求极值-搜答案-智慧作业帮

原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=（1x-y-yx-y）•xyy-1=1-yx-y•xyy-1=-xyx-y．故答案是：-xyx-y．

2x²+2xy+y²-4x+z-2倍根号z-3+2=0可化为x²+2xy+y²+x²-4x+4+(z-3)-2倍根号(z-3)+1=0即(x+y)&#

答：x+y+z=3y=2zy≠0,则z≠0所以：y=2z/3x+2z/3+z=2zx=z/3令z=3k,y=2k,x=k(xy+yz+zx)/(x²+y²+z²)=(2k

由题得x^2y^4=(xy^2)^2=(3/2)^2=9/4

x的平方+2xy+2y的平方-6y+9=0(x+y)^2+(y-3)^2=0y=3x=-y=-3x方-y方=0

令x/2=y/3=z/4=kx=2ky=3kz=4k(xy+yz+zx)/(5x^2+3y^2+z^2)=(2k*3k+3k*4k+4k*2k)/[5*(2k)^2+3*(3k)^2+(4k^2)]=

处理这类比例问题,有一个通用方法如果：x:y:z=a:b:c可以设x=aky=bkz=ck带入计算,就行了自己来试试吧~

x²+2xy+4y²+x²y²+1=8xy(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²

x-y=2两边同时平方又x^2+y^2=12所以xy=4原式=xy(x^2-2xy+y^2)=xy(x-y)^2=16

解；z(x)=2x+2y²z(y)=4xy+12y²dz=(2x+2y²)dx+(4xy+12y²)dy

xy的平方·(-x的平方y的平方)·(2分之1xy的三次方)的平方=(xy)^2*(-x^2y^2)*(xy)^6/4=-(xy)^10/4=-[(xy)^2]^5/4=-2^5/4=-8

原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y，由已知得（3x-2y）（x+y）=0，因为x+y≠0，所以3x

（X+Y+Z)²=X²+Y²+Z²+2(XY+YZ+XZ)X²+Y²+Z²=10²-2×8=84

X/3=Y/1=Z/2得X=3YZ=2Y代入XY+YX+XZ=99得Y方=9X方=9Y方=81Z方=4Y方=36最后2X81+12X9+9X36=594

设x/3=y/4=z/5=t则x=3t,y=4t,z=5t∴(xy+yz+zx)/(x²+y²+z²)=(12t²+20t²+15t²)/(

3[-(x+y)+2xy²-z]-2[(x+y)-xy²+z]-5[-3(x+y)-z]=3(-x-y+2xy²-z)-2(x+y-xy²+z)-5(-3x-3

3x-4y＝z,2x+y＝8z,解得：x=3z,y=2zxy+yz分之x二次方+y二次方-z二次方=(x^2+y^2-z^2)/(xy+yz)=(9z^2+4z^2-z^2)/(6z^2+2z^2)=

由题意得｛x+y+z=02x-y-7z=0=｛x+z=-yβ2x-7z=ypβ+p得x=2z,把x=2z带入β=-3z当x=2z,β=-3z时原式=【3×2z×（-3z）-（4z）²】除以（

设x/4=y/3=z/2=k【k不等于0】则x=4k,y=3k,z=2k把x、y、z的值代入原式中得到（4k）方+（3k）方/(12k方-8k方）化简得25k方/4k方,k方约分掉,可得到25/4【2

解方程组：｛2x-3y-z=0.(1)｛x+3y-14z=0.(2)(1)+(2)得：3x-15z=0即：x=5z,代入（1）式得y=3z所以：(4x²-5xy+z²)/(xy+y