-4xˇ2ˇ·(1 2xy-yˇ2)-3x·(xyˇ2-2xˇ2y)

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-4xˇ2ˇ·(1 2xy-yˇ2)-3x·(xyˇ2-2xˇ2y)
先化简,再求值:3xy²(-x²y+2x³y²)-4xy(-xy²)·

原式=-3x³y³+6x^4y^4+4x²y³*2x²y=-3x³y³+6x^4y^4+8x^4y^4=-3x³y&su

(xy-x^2)乘以(xy)/(x-y)

对.前提是x不等于y

已知xy=-2,x+y=3,求代数式(4xy+12y)+[7x-(3xy+4y-x)]的值

(4xy+12y)+(7x-(3xy+4Y-x))=4xy+12y+(7x-3xy-4y+x)=4xy+12y+7x-3xy-4y+x=(4-3)xy+(12-4)y+(7+1)x=xy+8y+8x当

(-3x^y+2xy)-( )=4x^+xy

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

已知x²+y²-2x+4y+5=0,求X^4-y^4/2x^2+xy-y²·2x-y/xy

x²+y²-2x+4y+5=0配方:(x-1)²+(y+2)²=0所以只有x-1=0且y+2=0∴x=1,y=-2∴(X^4-y^4)/(2x^2+xy-y&#

计算:4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2

4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2=(4xy-x²-2xy-y²)/[xy(x+y)(x-y)]÷[(x-y)(x+2y

x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2

x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2=16x^3y^3-49x^3y^3-45^3y^3=-78x^3y^3如果本题有什么不明白可以追问,

已知实数x,y满足xˇ2+yˇ2=1则(1-xy)(1+xy)的最大值

1^2=(x^2+y^2)^2=x^4+2x^2y^2+y^4>=2x^2y^2+2x^2y^2所以x^2y^2=00

(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)

(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)=1.=3(a-b)/(10ab)*(25a^2b^3)/((a+b)(a-b))=15/2*(ab^2)/(a+

Xˇ2-Xˇ2分之Xˇ2-XY+Yˇ2

分子、分母都除以X^2=-7/8

xy=-2,x+y=3,求代数式(4xy+12y)+[7x-(3xy+4y-x)]

(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

已知XY=8,满足Xˇ2Yˇ-XYˇ2-X+Y=56,求Xˇ2+Yˇ2的值

在线数学帮助你:这是一道关于化简求解的题目;根据题目给出的式子:X^2Y-XY^2-X+Y=56;提出XY;XY(X-Y)-(X-Y)=56;再提出(X-Y)可得;(XY-1)(X-Y)=56;因为X

x,y 属于R且xˇ2+4yˇ2=4,求xy的最大最小值

x^2+4y^2=4x^2+4y^2+4xy=(x+2y)^2=4+4xy≥0,xy≥-1x^2+4y^2-4xy=(x-2y)^2=4-4xy≥0,xy≤1xy的最大最小值分别是1,-1

已知x-y=4xy,则2x+3xy-2yx-2xy-y

∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.

xˇ+yˇ+2xy=256

(x+y)²=16²→x+y=16……①x+y=-16……②(x-y)²=2²→x-y=2……③x-y=-2……④①③联立解得:x1=9,y1=7①④联立解得:

已知x^2+y^2-4x+6y+13=0求(xy+y^2/5x^2-5xy+xy+y^2)·(5/x+y)-(y/x-y

x^2+y^2-4x+6y+13=0可以化成(x-2)^2+(y-3)^2=0所以x=2,y=3后面的代数你自己算吧

4x^2y+12xy+9y

4x^2y+12xy+9y=y(4x^2+12x+9)=y(2x+3)^2

x^2y+4xy+4y

x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²

先化简,再求值 4(xy²+3xy-2x²y)-[12xy-3(3x²y-2xy²

化简的结果是xy(x-2y)再问:过程?!再答:过程太麻烦了,电脑没法打。

已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]

x^2+y^2-2x+4y+5=0(x-1)^2+(y+2)^2=0x=1,y=-2化解式子,再代入