在已知图x² y²-4x 2y-4=0中,长为2的弦的中点的轨迹方程.

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已知x+y=-5,xy=7,求x2y+xy2-x-y的值.

x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.

若x+y=2,xy=-4,求x2y+xy2+1的值

(x+y)(xy)=x^2y+xy^2=-8原式=-7

2(x2y+xy)-3(x2y+xy)-4x2y其中x=-2,y=12

原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.

已知x+y=6,xy=4,则x2y+xy2的值为______.

∵x+y=6,xy=4,∴x2y+xy2=xy(x+y)=4×6=24.故答案为:24.

已知x-y=1,求代数式x4-xy3-x3y-3x2y+3xy2+y4.

原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(

已知x+y=10,xy=24,求x3+y3-x2y-xy2的值

x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10

一个多项式加上2x2y-3xy2-2x+1的2倍等于4x2y+5xy2+3x-2y+5,求这个多项式.

原式=(4x2y+5xy2+3x-2y+5)-2(2x2y-3xy2-2x+1)=4x2y+5xy2+3x-2y+5-4x2y+6xy2+4x-2=11xy2+7x-2y+3.

先化简后求值:4x2y-[6xy-3(4xy-2)-x2y]+1,其中x=2,y=-12

原式=4x2y-6xy+3(4xy-2)+x2y+1=5x2y+6xy-5当x=2,y=-12时,原式=5×4×(-12)+6×2×(-12)-5=-21.

已知x+y=6,xy=-3,则x2y+xy2=

那个2是平方吧?可以用^代替原式=x^y+xy^=xy(x+y)=-3*6=-18

当x=2011,y=2012时,求代数式3x3-4x3y2+3x2y+2x2+4x3y2+2x2y-5x2-5x2y+x

化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3

如果2x+y=4,xy=3,那么2x2y+xy2的值为______.

∵2x+y=4,xy=3,∴2x2y+xy2=xy(2x+y)=3×4=12.故答案为:12

已知A=x3+3x2y-5xy2+6y3-1,B=y3+2xy2+x2y-2x3+2,C=x3-4x2y+3xy2-7y

A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2

先化简,再求值:x2y-[4x2y-(xyz-x2z)-3x2z]-2xyx,其中x的倒数等于其本身,|y|=3,x2=

x=±1,y=±3,z=±2xyzz>y则0>x>z>yx=-1,y=-3,z=-2,x2y-[4x2y-(xyz-x2z)-3x2z]-2xyx=x2y-4x2y+xyz-x2z+3x2z-2xyx

已知X2+Y2+4=2X+XY+2Y,则X2Y的值是多少?

由题意得(x-2)平方+(y-2)平方+(x-y)平方=0,故x=y=2,故x平方y=8

化简求值:2(x2y+xy)-3(x2y-xy)-4x2y,其中x=-1,y=1.

原式=2x2y+2xy-3x2y+3xy-4x2y=-5x2y+5xy,当x=-1,y=1时,原式=-5×(-1)2×1+5×(-1)×1=-5-5=-10.

在matlab中怎样已知f(x,y)=sin(x2y)e-x-y,求d2f/dxdy

楼上兄的回答思路是正确的,只不过修正一下小错误symsxyf=sin(x^2*y)*exp(-x-y);ddf=diff(diff(f,x),y);simple(ddf)

关于x,y的方程组 3x+2y=m+1,4x2y=m-1求y,x

如果x,y符号相反,绝对值相等,即y=-x,代入原方程组,得3x-2x=m+1,4x-2x=m-1,即x=m+1,2x=m-1解之,2(m+1)=m-1,得m=-3如果x比y大1,即x=y+1,代入原

已知X2+4y2-4x+4y+5=0 求x-y的值 已知xy=4满足x2y-xy2-x+y=56,求x2+y2的值

x²+4y²-4x+4y+5=0(x-2)²+(2y+1)²=0x-2=0x=22y+1=0y=-1/2x-y=2+1/2=5/2x²y-xy

已知x+2y=5,xy=1.则2x2y+4xy2=______.

∵x+2y=5,xy=1,∴2x2y+4xy2=2xy(x+2y)=2×1×5=10,故答案为:10.