3道分式的运算题目

1、
(1/x)+(1/y)
= (x+y)/(xy)
= 1/(x+y)
所以：(x+y)^2 = xy（注：这道题的x和y要涉及到复数,在实数范围内(x+y)^2 = xy无解）
即：x^2 + y^2 = -xy
所以：(y/x)+(x/y) = (x^2+y^2)/xy = -xy/xy = -1
2、
X-[2/(X+1)]
= (x^2+x-2) / (x+1)
= [ (x-1)(x+2) ] / (x+1)
所以：
[(x^2+2x)/(x+1)]÷{X-[2/(X+1)]}
= [ x(x+2)/(x+1) ] × { (x+1)/[(x-1)(x+2)] }
= x/(x-1)
3、
（1）
x^2 - 4x + 1 = 0
显然x不等于0,两边同除以x,得：
x - 4 + (1/x) = 0
即：x + (1/x) = 4
[ x + (1/x) ]^2 = x^2 + 2 +(1/x^2) = 4^2 = 16
所以：x^2 + (1/x^2) = 16-2 = 14
（2）
先求倒数：
(x^4+x^2+1)/x^2
= x^2 + 1 + (1/x^2)
= 14 + 1 = 15
所以：
x^2/(x^4+x^2+1) = 1/15