若方程组3x 2y=m 1,4x 3y=m-1的解满足xy>0且m为整数

原式=x3-2y3-3x2y-3x3+3y3+7x2y=-2x3+y3+4x2y

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．

原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2，y=12时，原式=-9．

∵代数式x3+y3+3x2y+axy2含有因式x-y，∴当x=y时，x3+y3+3x2y+axy2=0，∴令x=y，即x3+x3+3x3+ax3=0，则有5+a=0，解得a=-5．将a=-5代入x3+

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．

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

（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

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

根据题意得：（x3-3x2y）-（3x2y-3xy2）=x3-3x2y-3x2y+3xy2=x3-6x2y+3xy2，故选C．

原式=x3+3x2y-5xy2+6x3+1-2x3+y3+2xy2+x2y+2-4x2y-7x3-y3+4xy2+1=-2x3+xy2+4，由于y为偶次幂，故误把“x=3，y=-1”写成“x=3，y=

（1）（x3-2x2y+3y2）-（-2x3-3x2y+5y2）=x3-2x2y+3y2+2x3+3x2y-5y2=3x3+x2y-2y2，答：这个多项式为3x3+x2y-2y2．（2）当x=-12，

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

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

5x1+4x2+3x3=(3x1+x2+x3)+(2x1+3x2+2x3)≤840+700=1540所以最大值为1540

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

由题意得：多项式3x2y-4xy2+x3-5y3按y的降幂排列是-5y3-4xy2+3x2y+x3．故答案是：-5y3-4xy2+3x2y+x3．

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

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．

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

x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)