已知x,y的值满足方程组x+2y=3,x-y=-2,求x/(x-y)²+x²-y²/(x+y)²的值
问题描述：
已知x,y的值满足方程组x+2y=3,x-y=-2,求x/(x-y)²+x²-y²/(x+y)²的值
答
由x+2y=3,x-y=-2得x＝-1／3 y＝5／3
x+2y=3 x+y=3-y=4/3
x/(x-y)²+(x²-y²)/(x+y)²
=x/(x-y)²+(x+y)(x-y)/(x+y)²
=x/(x-y)²+(x-y)/(x+y)
=(-1/3)/(-2)2-2/(4/3)
=-1/12-3/2
=-19/12
答
x+2y=3 (1)
x-y=-2 (2)
消去y,(1)+(2)×2得：
x+2x=3-4
3x=-1
x=-1/3
消去x,(1)-(2)得：
2y+y=3+2
3y=5
y=5/3
方程组的解为 x=-1/3,y=5/3
x/(x-y)²+(x²-y²)/(x+y)²
=x/(x-y)²+(x+y)(x-y)/(x+y)²
=x/(x-y)²+(x-y)/(x+y)
=(-1/3)/(-1/3-5/3)²+(-1/3-5/3)/(-1/3+5/3)
=(-1/3)/(-2)²+(-2)/(4/3)
=(-1/3)/4-2×3/4
=-1/12-3/2
=-1/12-18/12
=-19/12