[3/(x-2)+2/(x+2)]÷(5x²+2x)/(x²-4)
=[3/(x-2)+2/(x+2)]*(x²-4)/(5x²+2x)
=[3/(x-2)+2/(x+2)]*(x-2)(x+2)/(5x²+2x)
=3/(x-2)*(x-2)(x+2)/(5x²+2x)+2/(x+2)*(x-2)(x+2)/(5x²+2x)
=3(x+2)/(5x²+2x)+2(x-2)/(5x²+2x)
=[3(x+2)+2(x-2)]/(5x²+2x)
=[3x+6+2x-4]/(5x²+2x)
=[5x+2]/[x(5x+2)]
=1/x
=1/3其中x=3 呢？？？不是有了吗？不是先算（3/x-2+2/x+2）÷5x²+2x/x²-4， 得数算出后 x=3再代入.真的有吗-我找不到..化简得[3/(x-2)+2/(x+2)]÷(5x²+2x)/(x²-4)=1/x 代入x=3=1/3