接关羽x,y的方程组x 2y=m 5 3x-4y=4m-15

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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

解方程组2X+3Y=15M,5X-3Y=-M,X,Y的值

2X+3Y=15M,5X-3Y=-M两式相加7X=14MX=2MY=11/3M

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=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

如果(3x2y-2xy2)÷m=-3x+2y,则单项式m为(  )

根据题意得:(3x2y-2xy2)÷(-3x+2y)=-xy,则m=-xy.故选B.

x+y=5,xy=2,求代数式-x2y-xy2的值

解-x²y-xy²=-xy(x+y)=-2×5=-10

方程组2x+y=1+3m,x+2y=1-m的解满足X+Y

2x+y=1+3m(1)x+2y=1-m(2)(1)+(2)得:3x+3y=2+2m所以:x+y=(2+2m)/3因为x+y〈0则:(2+2m)/3

(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

若x+y=5,xy=6,则x2y+xy2的值为______.

∵x+y=5,xy=6,∴x2y+xy2=xy(x+y)=5×6=30.故答案为:30.

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

数学竞赛题:若实数x,y满足方程组xy+x+y+7=0,3x+3y=9+2xy,则x2y+xy2=?

x2y+xy2=xy*(x+y)因为x+y=-(7+xy)又x+y=(9+2xy)\3所以(9+2xy)\3=-(7+xy)3+2xy\3=-7-xy5xy\3=-10解得xy=-6所以x+y=-(7

已知关于x,y的方程组{x+2y=3m,x-y=9m

x的值比y的值小5y-x=5x-y=-5x-y=9m=-5m=-5/9x+2y=3m(1)x-y=9m(2)(1)*5+(2)*45(x+2y)+4(x-y)=5*3m+4*9m9x+6y=51m3x

当x=-1,y=1时求代数式2x2y-(5xy2-3x2y)-x2的值

代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方

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

若x-y=3,xy=-2,则xy2-x2y的值是______.

原式=-xy(x-y),当x-y=3,xy=-2时,则原式=-3×(-2)=6.故答案为:6.

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