设x y-xz z-2=0则dz dx

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已知x、y是实数且满足x2+xy+y2-2=0,设M=x2-xy+y2,则M的取值范围是______.

由x2+xy+y2-2=0得:x2+2xy+y2-2-xy=0,即(x+y)2=2+xy≥0,所以xy≥-2;由x2+xy+y2-2=0得:x2-2xy+y2-2+3xy=0,即(x-y)2=2-3x

设siny-e^x+xy^2=0,求dy/dx

siny-e^x+xy^2=0cosy.y'-e^x+2xy.y'+y^2=0(cosy+2xy)y'=e^x-y^2y'=(e^x-y^2)/(cosy+2xy)

已知X.Y是实数且满足X^2+XY+Y^2-2=0,设M=X^2-XY+Y^2,则M的取值范围是

X^2+XY+Y^2-2=0X^2+XY+Y^2=2x^2+y^2=2-xy又:(x-y)^2≥0xy≤(x^2+y^2)/2=(2-xy)/2=1-xy/2(1+1/2)xy≤1xy≤2/3∴M=X

设X~N(0,1),Χ^2(5),XY相互独立,令Z=X/Y/5则Z=

第一个无过程,就是考察t分布的定义,这里结果是t(5);第二个也可以说是无过程,考察的是二项分布的数字特征及矩估计方法(替换原理)这两个常识.对于X服从B(n,p)来说,其期望为EX=np,方差为DX

高数题,设z=x^2+xy+y^2,则dz=

dz=2x+y就是对z求x的导数吧

设xy>0,且xy=4x+y+12,求xy的最小值

xy-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以

设xy-xe^(y^2)-x=-2,则当x=1,y=0时,dy/dx=

y+xdy/dx-e^(y^2)-2xe^(y^2)dy/dx-1=0x=1,y=0dy/dx-1-2dy/dx-1=0dy/dx=-2

设实数xy满足X平方+Y平方-2Y=0,则X平方+Y平方的最大值是

化为三角函数x=costy=sint+1x^2+y^2=(cost)^2+(sint+1)^2=2sint+2最大为4

设函数Y=f(x)由方程xy+y^2-2x=0,则dy/dx=?

xy+y^2-2x=0y+xy'+2yy'-2=0(x+2y)y'=2-yy'=(2-y)/(x+2y)dy/dx=(2-y)/(x+2y)

设实数X,Y满足X^2+2XY-1=0,则XY的取值范围是?这个是对的

由题得xy=(1-x2)/2,因为x2>=0,所以xy=(1-x2)/2

设sin(xy)-x^2y=0,求dy/dx,求导什么的都要疯了!

等式两边对x求导:cos(xy)*(y+x*y')-(2x*2y+x^2*2*y'=0解出y'即为所求

设x>0,y>0,x+y+xy=2,则x+y的最小值是(  )

∵x>0,y>0,∴x+y≥2xy(当且仅当x=y时取等号),则xy≤x+y2,xy≤(x+y)24,∵x+y+xy=2,∴xy=-(x+y)+2≤(x+y)24,设t=x+y,则t>0,代入上式得,

设e^xy-xy^2=Siny,求dy/dx

你好!两边对x求导:e^(xy)*(y+xy')-y^2=y'cosy解得y'=(y^2-ye^(xy))/(xe^(xy)-cosy)

设实数xy满足x2 +2y2=6,则x+y的取值范围

解由x2+2y2=6得x2/6+y2/3=1故设x=√6cosa,y=√3sina则x+y=√6cosa+√3sina=3(√6/3cosa+√3/3sina)=3sin(a+θ)由-3≤3sin(a

设A=2x²-3xy.

|x-2a|+(y+3)²=0所以x-2a=0,y+3=0x=2a,y=-3B-2A=(4x²-6xy+2y²-3x-y)-2(2x²-3xy+y²-

设f(x+y,xy)=x^2+y^2,则f(x,y)

f(x+y,xy)=x^2+y^2=(x+y)^2-2xyf(x,y)=x^2-2y

定义A⊗B={z|z=xy+xy,x∈A,y∈B}.设集合A={0,2},B={1,2},C={1}.则集合(A⊗B)⊗

由题意可求(A⊗B)中所含的元素有0,4,5,则(A⊗B)⊗C中所含的元素有0,8,10,故所有元素之和为18.故选C

设Z=x²+2xy,求dz

z=x^2+2xy两边同时求导数,得到:dz=2xdx+2ydx+2xdy即:dz=2(x+y)dx+2xdy.