x² y² z²≤2z z≤x² y²所确定立体的体积
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∑是循环和例如∑a=a+b+c∑a^2=a^2+b^2+c^2∑(z-y)(x-y)/(x+y-2z)(y+z-2x)=∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z
令x/3=y/4=z/5=k;则x=3k,y=4k,z=5k;(xy+yz+xz)/(xx+yy+zz)=(3k*4k+4k*5k+3k*5k)/(3k*3k+4k*4k+5k*5k)=(12+20+
X:Y:Z=1:2:3因为:14(XX+YY+ZZ)=(X+2Y+3Z)^214(XX+YY+ZZ)-(X+2Y+3Z)^2=013X^2+10Y^2+5Z^2-4XY-6XZ-12YZ=0(4X^2
设z=x+yi,x、y∈R,由于zz-1=x+yix-1+yi=(x+yi)(x-1-yi)(x-1+yi)(x-1-yi)=x2+y2-x(x-1)2+y2+y(x-1)2+y2i是纯虚数,故有x2
有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y
x-z=(x-y)+(y-z)=5+4=9(x-y)²=x²-2xy+y²=25(y-z)²=y²-2yz+z²=16(x-z)²
∵x-2y+z=(x-y)-(y-z),x+y-2z=(y-z)-(z-x),y+z-2x=(z-x)-(x-y).设x-y=a,y-z=b,z-x=c,则原式=-ac/(a-b)(b-c)+(-ba
(X+Y+Z)*(X+Y+Z)=XX+YY+ZZ+2(XY+YZ+XZ)=1,又XY+YZ+XZ=0,所以XX+YY+ZZ=1
正整数?取对数即证:2xlnx+2ylny+2zlnz>(y+z)lnx+(x+z)lny+(x+y)lnzx>y>z,lnx>lny>lnz由排序不等式得xlnx+ylny+zlnz>ylnx+zl
X=3K,Y=K,Z=2K(2X方-2Y方+5Z方)/(XY+YZ+ZX)=(2*9-2+5*4)/(3+2+6)=36/11
由于f'(x)=arcsiny+2xz则f“(xz)=2x;同理,f'(y)=x/√(1-y²)+z²则f"(yz)=2z;f'(z)=2yz+x²则f"(zz)=2y
xx'+yy'+zz'可以由前面的推导过来.|a|=√(x^2+y^2+z^2)|b|=√(x'^2+y'^2+z'^2)|c|=√((x-x')^2+(y-y')^2+(z-z')^2)cosθ=|
x2+y2-z2+2xy/x2-y2+z2-2xz=(x+y)2-z2/(x-z)2-y2=(x+y-z)(x+y+z)/(x-y-z)(x-z+y)=(x+y+z)/(x-y-z)然后就是代入了
用一个变量来表示另外两个变量即可得解.如用Y表示X有X=3Y/4用Y表示Z有z=5Y/4所以(xy+yz+xz)/xx+yy+zz=(3YY/4+5YY/4+15YY/16)/(9YY/16+YY+2
这是线性规划的题目根据x-y≥-1,x+y≥1,2x-y≤1画图可以发现满足的点位于三条直线所围成的三角形内.x-y=-1与2x-y=1交点为x=2,y=3即交点为(2,3)Z=(x-2y)/(x+y
如果是xy+xz+yz的话:xy+xz+yz=[(x+y+z)^2-(x*x+y*y+z*z)]/2=5*5-6=19
x=1,y=0,z=9首先x、y、z都是个位数xyy可以写成100x+10y+y同理,zz可以写成10z+zyx写成10y+x等式重新代入以上化解后的式子,就是:100x+11y-11z=10y+x合
第二个分母写错了?(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)=1
(x+y)^2=(x-y)^2+4xy=64+4(-z^2-16)=-4z^2=0所以(x+y)^2=0所以x+y=0x-y=8x=4,y=-4z=0