3道分式的运算题目1.如果分式(1/x)+(1/y)=1/(x+y),那么分式(y/x)+(x/y)的值为( )A.1

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 = -12、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