x y 2=12 3x y=22

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x y 2=12 3x y=22
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-1|+|y+3|=0,则1-xy-xy2 =?

|x-1|+|y+3|=0,有|x-1|≥0|y+3|≥0所以必须有|x-1|=0|y+3|=0才可以满足所以x=1y=-3代入1-xy-xy²=1+3-9=-5

若x+y=2,xy=-4,求x2y+xy2+1的值

(x+y)(xy)=x^2y+xy^2=-8原式=-7

已知A=x3-2y3+3x2y+xy2-3xy+4,B=y3-x3-4x2y-3xy-3xy2+3,C=y3+x2y+2

因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.

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

若|x-1|+(y+3)2=0,求1-xy-xy2的值.

由题意得,x-1=0,y+3=0,解得x=1,y=-3,所以,1-xy-xy2=1-1×(-3)-1×(-3)2,=1+3-9,=4-9,=-5.

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

2(xy-5xy2)-(3xy2-xy)=(2xy-10xy2)-(3xy2-xy)=2xy-10xy2-3xy2+xy=(2xy+xy)+(-3xy2-10xy2)=3xy-13xy2,∵(x+1)

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

有这样一道题:“计算(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)的值,其中x=1

(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的

计算:3xy(x2y-xy2+xy)-xy2(2x2-3xy+2x)因式分解

3xy(x²y-xy²+xy)-xy²(2x²-3xy+2x)=3x³y²-3x²y³+3x²y²-

当x+y=5 xy=6 (x2y)(xy2)=?

(x2y)(xy2)=xy*xy*xy=6*6*6=216楼主是不是抄错题了?(x2y)+(xy2)=xy(x+y)=5*6=30

已知xy=-1,求3xy2(x-x3y2-12

3xy2(x-x3y2-12x2y)=3x2y2-3x4y4-32x3y3,当xy=-1时,原式=3×(-1)2-3×(-1)4-32×(-1)3=32.

若x2y+xy2=30,xy=6,求下列代数式的值:(1)x2+y2;(2)x-y.

因为x^2y+xy^2=30,xy(x+y)=30,xy=6x+y=5所以(x+y)^2=x^2+y^2+2xy=x^2+y^2+12=25所以x^2+y^2=13所以(x-y)^2=x^2+y^2-

化简并求值:(2x2y-2xy2)-[(-3x2y2+3x2y)+(3x2y2-3xy2)],其中x=−12,y=2

原式=2x2y-2xy2+3x2y2-3x2y-3x2y2+3xy2=-x2y+xy2,当x=-12,y=2时,原式=-(−12)2×2+(-12)×22=-52.

已知A=8x2y-6xy2-3xy,B=7xy2-2xy+5x2y,若A+B-3C=0,求C-A.

由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6

当x等于1,y等于2时,求代数式3xy2-【2xy2-2(xy-1.5x2y)】+xy-3x2y的值

解3xy²-[2xy²-2(xy-1.5x²y)]+xy-3x²y=3xy²-(2xy²-2xy+3x²y)+xy-3x²

因式分解x3+x2+xy-xy2

x^3+x^2+xy-xy^2=x(x^2+x+y-y^2)=x(x^2-y^2+x+y)=x[(x+y)(x-y)+(x+y)]=x(x+y)(x-y+1)