题目

分式综合题
1.计算：
(1-x)/(x+2)²÷(x-2+(3/x+2))
2.已知x²+y²-4x+6y+13=0,
求(x²-4xy+4y²)/(x²-4y²)的值

提问时间：2021-03-17

答案

[(1-x)/(x+2)^2]/[x-2+3/(x+2)]
=[(1-x)/(x+2)^2]/{[(x-2)(x+2)+3]/(x+2)}
=(1-x)(x+2)/[(x+2)^2(x^2-4+3)]
=-(x-1)/[(x+2)(x+1)(x-1)]
=-1/[(x+1)(x+2)]
=-1/(x^2+3x+2)
x^2+y^2-4x+6y+13=0,
(x-4x+4)+(y^2+6x+9)=0
(x-2)^2+(y+3)^2=0
平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立
所以两个都等于0
所以x-2=0,y+3=0
x=2,y=-3
(x^2-4xy+4y^2)/(x^2-4y^2)
=(x-2y)^2/(x+2y)(x-2y)
=(x-2y)/(x+2y)
=(2+6)/(2-6)
=-2