方程4x²-y² 4x 2y=0表示的曲线是

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若x+y=2,xy=-4,求x2y+xy2+1的值

(x+y)(xy)=x^2y+xy^2=-8原式=-7

先化简,再求值(3x2y-2xy2)-(xy2-2x2y),其中x=-1,y=2.

(3x2y-2xy2)-(xy2-2x2y)=3x2y-2xy2-xy2+2x2y=5x2y-3xy2当x=-1,y=2时,原式=5×(-1)2×2-3×(-1)×22=10+12=22.

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.

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.

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

(X+Y)2=1402X2Y*3=14400

(X+Y)2=1402X2Y*3=14400(X+Y)2=140→X+Y=70→Y=70-X①2X2Y*3=14400→XY=1200②把①代人②得:X(70-X)=1200X²-70X+1

求微分方程的通解(xy2-x)dx+(x2y+y)dy=0

(xy2-x)dx+(x2y+y)dy=0y(x²+1)dy=-x(y²-1)dxy/(y²-1)dy=-x/(x²+1)dx两边积分得ln|y²-1

(-2x2y)3*4x-3sup>=

题目1看不明白解题目2x+y=4,(x+y)^2=4^2=16,同样x-y=10,(x-y)^2=10^2=100,(x+y)^2=x^2+2xy+y^2,(x-y)^2=x^2-2xy+y^2,(x

已知(x-2)2+|y+1|=0,求5xy2-[2x2y-(3xy2-2x2y)]的值.

原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.

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.

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.

已知x-y≠0 x2-x=7 y2-y=7 求x3+y3+x2y+xy2的值

x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup