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已知:x-y=1,求证:x3-3xy-y3=

解:x^3+3xy+y^3=(x+y)(x^2-xy+y^2)+3xy, =(x^2-xy+y^2)+3xy, =(x+y)^2-3xy+3xy, =1.

y³+x³-3xy=1 在方程两端分别对 x 求导,得 3y²y'+3x²-3(y+xy')=0 y²y'+x²-y-xy'=0 从而 y'=(y-x²)/(y²-x)

已知x+y=1,求x³+y³+3xy的值 x³+y³+3xy =(x+y)(x²-xy+y²)+3xy =x²-xy+y²+3xy =x²+2xy+y² =(x+y)² =1

∵x=12+3=2-3,y=12?3=2+3,∴xy=1,x+y=4,∴x3+12xy+y3=12+x3+y3,=12+(x+y)(x2-xy+y2),=12+(x+y)[(x+y)2-3xy],=12+4×(42-3×1),=64.故本题答案为64.

x+y=3+52+3?52=3x?y=3+52?3?52=9?54=1∴x3+y3=(x+y)(x2-xy+y2)=(x+y)[(x2+2xy+y2)-3xy]=(x+y)?[(x+y)2?3xy]=3?(32?3)=18,故答案为18.

因为xy为正整数,所以xy+61大于0,所以x大于y,根据整数具有离散性,所以x大于等于y+1,所以x-1大于等于y,所以将x3-y3=xy+61转化得x3=xy+61+y3,放缩为x3

解:将方程变形为:x^3+3xy(x+y)+y^3=1^3,刚好是(x+y=1)的三次方展开式。经分解,原方程可变为:(x+y-1)(x^2-xy+y^2+x+y+1)=0。∴原方程曲线由直线y=1-x和椭圆x^2-xy+y^2+x+y+1=0(∵根据δ

x³-y³=(x-y)(x²+xy+y³)=218 x-y=2 x=y+2 带入1式,(y+2)²+(y+2)y+y²=109 y²+2y-35=0 y=5 y=-7 x=7 x=-5 所以两组解,x=7,y=5 x=-5,y=-7

(x+y)3-x3-y3+3xy =x³+3x²y+3xy²+y³-x³-y³+3xy =3x²y+3xy²+3xy =3xy(x+y+1)

2x^4y^4-X^3Y^4=(xy)^3(2x-y)= (1/3)^3(-3)= -1/9

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