y2-2xy+9=0的隐函数

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y2-2xy+9=0的隐函数
已知x2-xy=-3,2xy-y2=-8,求2x2+4xy-3y2的值.

x2-xy=-3①,2xy-y2=-8②,①×2+②×3得:2x2-2xy+6xy-3y2=-6-24=-30,则2x2+4xy-3y2=-30.

求所确定的隐函数的导数dy/dx y2-2xy+9=0 不明白方程两边分别对x求导数怎么出来的

y^2-2xy+9=0两边分别对x求导数得:2y(dy/dx)-2y-2x(dy/dx)=0(dy/dx)(y-x)-y=0dy/dx=y/(y-x)

25(x2+ 2xy +y2)-9(x2-2xy+y2)因式分解.x2,y2是x平方,y平方的意思.2xy就是2xy.快

25(x²+2xy+y²)-9(x²-2xy+y²)=[5(x+y)]²-[3(x-y)]²=[5(x+y)+3(x-y)][5(x+y)-

已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?

已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?2x-3*根号(xy)-2y=0(根号X-2根号Y)(2根号X+根号Y)=0根号X-2根号Y

设x≥0,y≥0,且x+2y=1,求函数y=log1/2(8xy+4y2+1)的值域

x=1-2y,所以8xy+4y²+1=8(1-2y)×y﹢4y²+1=﹣12y²+8y+1y在[0,0.5]对称轴是1/3所以y=⅓时,最大值为7/3y=0时

已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值

已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*

因式分解:25(x2+2xy+y2)-9(x2-2xy+y2)

原式=25(x+y)²-9(x-y)²=[5(x+y)+3(x-y)][5(x+y)-3(x-y)]=(8x+2y)(2x+8y)=4(4x+y)(x+4y)

若X2+Y2-2X-6Y+10=0 ,求(x2-y2)/xy的值

10拆成1+9X2-2X+1+Y2-6Y+9=0(X-1)2+(Y-3)2=0平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以X-1=0,Y-3=0X=1,Y=

已知X2-2x+y2+6y+10=0,求(x2-2xy)/(xy+y2)的值

x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3∴(x2-2xy)/(xy+y2)=(1²-2*(-3))/

已知2x2-3xy+y2=0(xy≠0),则xy+yx的值是(  )

根据题意,2x2-3xy+y2=0,且xy≠0,故有(yx)2−3yx+2=0,即(yx−1)(yx−2)=0,即得yx=1或2,故xy=1或12,所以xy+yx=2或212.故选A.

设x》0,y》0,且x+2y=1,求函数y=log1/2的(8xy+4y2+1)次方的值域

x>0,y>0,x+2y=1,所以x=1-2y.因为1-2y>0,所以0再问:是大于等于0再答:那么值域是[log1/27/3,0)】

设二元函数z=x2+xy+y2—x-y,x2+y2≤1,求它的最大值和最小值.

2z=2x^22xy2Y^2-2x-2y=(x^22xyy^2)(x^2-2x)(y^2-2y)2z2=(x^22xyy^2)(x^2-2x1)(y^2-2y1)=(xy)^2(x-1)^2(y-1)

已知xy=27,则x2−3xy+2y22x2−3xy+7y2的值是(  )

由xy=27得,x=27y,∴x2−3xy+2y22x2−3xy+7y2=449y2−67y2+2y2849y2−67y2+7y2=60309=20103.故选C.

求函数Z=X2-XY+Y2-2X+Y的极值、、、步骤

换元.可设x=a+b,y=a-b.则z=2(a²+b²)-(a²-b²)-2(a+b)+(a-b)=a²-a+3b²-3b=[a-(1/2)

已知2x2+xy=6,3y2+2xy=9,求4x2+8xy+9y2的值为______.

∵4x2+8xy+9y2=4x2+2xy+9y2+6xy=2(2x2+xy)+3(3y2+2xy),2x2+xy=6,3y2+2xy=9,∴2(2x2+xy)+3(3y2+2xy)=2×6+3×9=3

求解方程x2+y2-xy=1确定的隐函数的导数dy/dx

太多了,只解一题:  13、分别计算左、右导数   f'(1-0)=lim(x→1-0)[f(x)-f(1)]/(x-1)=lim(x→1-0)[(x²+1)-2]/(x-1)=lim(x→

已知x2-y2=xy,且xy≠0,求代数式x2y-2+x-2y2的值.

∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a

分解因式:x2-2xy+y2-9

x2-2xy+y2-9,=x2-2xy+y2-9,=(x2-2xy+y2)-9,=(x-y)2-32,=(x-y+3)(x-y-3).

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²