(1)已知x^2+y^2=5 [1]
xy=2 [2]
[1]+[2]*2 (x+y)^2=9
x+y=±3
所以1/(1/x+1/y)＝xy/(x+y)=2/(±3)=±2/3
(2)计算:1/(a-x)-1/(a+x)-2x/(a^2+x^2)-(4x^3)/(a^4+x^4)+(8x^7)/(x^8-a^8).
=(a+x-a+x)/(a^2-x^2)-2x/(a^2+x^2)-(4x^3)/(a^4+x^4)+(8x^7)/(x^8-a^8).
=2x*(a^2+x^2-a^2+x^2)/(a^4-x^4)-(4x^3)/(a^4+x^4)+(8x^7)/(x^8-a^8).
=4x^3*(a^4+x^4-a^4+x^4)/(a^8-x^8)+(8x^7)/(x^8-a^8).
=8x^7/(a^8-x^8)-8x^7/(a^8-x^8)
=0