计算：(x^2+6x+10)/(x^2+6x+9)+(x^2-11)/(x^2-9)+(2x^2+8x+5)/(x^2+

看三个分母的因式可以知道：
首先,每个提出一个常数,把分子化成简单的常数,即化为：
1+1/(x^2+6x+9)+1-2/(x^2-9)+2-1/(x^2+4x+3)
接着,每个因式分母提出（x+3）,即为：
4+1/(x+3) * [1/(x+3)-2/(x-3)-1/(x+1)]
最后,中括号内通分化简即可：
4+1/(x+3) * [(x-3)(x+1)-2(x+3)(x-1)-(x+3)(x-3)/(x+3)(x-3)(x+1)]
=4+1/(x+3) * (x^2-2x-3-2x^2-4x+6-x^2+9)/[(x+3)(x-3)(x+1)]
=4+1/(x+3) * (-2x^2-6x+12)/[(x+3)(x-3)(x+1)]
=4-1/(x+3) * (2x^2+6x-12)/[(x+3)(x-3)(x+1)]
=4-2/(x+3) * (x-1)(x+4)/[(x+3)(x-3)(x+1)]