7x 4y-10=0和4x 2y-5=0

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4xy2-4x2y-y3(2和3是次方)

解题思路:提取公因式解题过程:提取公因式最终答案:略

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.

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.

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.

如果x+y=0,xy=-7,求①x2y+xy2;  ②x2+y2.

∵x+y=0,xy=-7,∴①x2y+xy2=xy(x+y)=-7×0=0;②x2+y2=(x+y)2-2xy=14.

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

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

(-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

4x2y-{x2y-「3xy2 – 1/ 2(4x2y-8xy2)+x2y」}-5xy2

答案:2x^2y+2xy^2原式=4x2y-{x2y-[3xy2-2x2y+4xy2+x2y]}-5xy2=4x2y-{x2y-[7xy2-x2y]}-5xy2=4x2y-{x2y-7xy+x2y]}

已知长方形的周长是28cm,相邻两边的长分别是xcm 和ycm,且 x3+x2y-xy2-y3=0,求长方形的面积。

解题思路:本题第一个方程容易列出,关键是由第二个条件,经过化简列出第二个方程。解题过程:

单项式-3x2减去单项式-4x2y,-5x2,2x2y的和,列算式为______,化简后的结果是______.

根据题意得,算式为:-3x2-[-4x2y+(-5x2)+2x2y],-3x2-[-4x2y+(-5x2)+2x2y]=-3x2+4x2y+5x2-2x2y=2x2+2x2y=2x2(1+y).

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

单项式5x2y、3x2y、-4x2y的和为______.

5x2y+3x2y+(-4x2y)=(5+3-4)x2y=4x2y,故答案为:4x2y.

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.

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

单项式5x2y,3x2y2,-4xy2的和为______.

5x2y+3x2y2+(-4xy2)=5x2y+3x2y2-4xy2.

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

如果x+y=0,xy=-7,x2y+xy2=______,x2+y2=______.

解;∵x+y=0,xy=-7∴x2y+xy2=xy(x+y)=-7×0=0x2+y2=(x+y)2-2xy=02-2×(-7)=0+14=14.