1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/(x+4)

1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1) - 1/(x+2)
同理
1/(x²+5x+6)=1/(x+2) - 1/(x+3)
1/(x²+7x+12)=1/(x+3) - 1/(x+4)
所以 方程成为 1/(x+1) - 1/(x+4) = 1/(x+4)
解得 x=2