再化简(x y)(x-2y)-my(nx-y)-搜答案-灵鹊做题网

对.前提是x不等于y

集合M={（x,y）|xy

因为x属于z,z代表整数又因为xy0,y

2m-6=-2m=2

m=x³-3x²y-2xy²+3y³

m=x³-3x²y+2xy²+3y³①n=x³-2x²y+xy²-5y³②①×3得3m=3x³-9x²

x+2y=1（1）x-2y=m（2）（1）+（2）得：2x=1+mx=（1+m）/2>11+m>2得：m>1（1）-（2）得：4y=1-my=（1-m）/4>=-11-m>=-4-m>=-5得：m

【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy

（x^2-y^2分之M）=(x^2-y^2分之2xy-y^2)+(x+y分之x-y）,M/(x²-y²)=(2xy-y²)/(x-y)(x+y)+(x-y)/(x+y)M

解M/(x²-y²)-(x-y)/(x+y)——是这个吗?=M/(x²-y²)-(x-y)²/(x²-y²)=[M-(x²

2x^3-7x^y+5xy^2+14y^3=3(x^3-3x^2y+2xy^2+3y^3)-(x^3-2x^2y+xy^2-5y^3)=3m-n

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

M=(2x^2y+xy^2)（x-2y)-(2x^2-3xy+xy^2)(-xy)很简单啊.

如果多项式只有二项,则M为2m-2=0m=2

点集m={(x,y)|xy

{(x,y)|xy

(1)X+Y=M(2)2X-Y=6(1)+(2)3X=M+6X=(M+6)/3=M/3+2Y=2X-6=(2M-6)/3=2M/3-2XY是整数所以M是3的倍数XY

由题意可得：(2xy-y²)/(x²-y²)+(x-y)/(x+y)=[(2xy-y²)+(x-y)²]/(x²-y²)=x

m/x²-y²=[2xy-y²+(x-y)²]/(x²-y²)=(2xy-y²+x²-2xy+y²)/(x&#

m³-2m²-4m+8=m²(m-2)-4(m-2)=(m-2)(m²-4)=(m+2)(m-2)²x²-2xy+y²-9=(x-

xy>0得x,y同号又因为x+y

3(1,4)和0(2,2)

1.X=(1+M)/2Y=(1-M)/42.M大于1小于等于5