1/(x^2+3x+2)+x^2/(x-4)=3/(x^2-x-2) +1

1/(x^2+3x+2)+2/(x^2-4)=3/(x^2-x-2) +1
1/[(x+2)(x+1)] + 2/[(x+2)(x-2)] = 3/[(x-2)(x+1)] +1
所以 x≠-1, x≠-2, x≠2
两边同时乘以 (x+2)(x+1)(x-2)
x-2+2(x+1)=3(x+2)+(x+2)(x+1)(x-2)
(x+2)(x+1)(x-2)+6=0
x^3+x^2-4x+2=0
x^3-x^2+2x^2-2x-2x+2=0
x^2(x-1)+2x(x-1)-2(x-1)=0
(x-1)(x^2+2x-2)=0
x=1, x=-1+√3, x=-1-√3
检验：
x=1时，左＝1/6-2/3=-1/2，右＝-3/2+1=-1/2，左＝右
x=-1+√3时，左＝(1-√3)/2，右＝(1-√3)/2，左＝右
x=-1-√3时，左＝(√3+1)/2，右＝(√3+1)/2，左＝右
所以：
原方程有3解：x=1, x=-1+√3, x=-1-√3

显然不正确，把x=0代入得
左边等于1/2，而右边等于-1/2

1/(x^2+3x+2)+x^2/(x-4)=3/(x^2-x-2) +1

1/((x+1)(x+2))+x^2/(x-4)=3/((x+1)(x-2))+1

1/(x+1)-1/(x+2)+x^2/(x-4)=1/（x-2）-1/(x+1)+1
你把分母都乘开 计算是麻烦点

是1/(x^2+3x+2)+x^2/(x^2-4)=3/(x^2-x-2) +1吧

方程可化为1/(x+1)(x+2)+x^2/(x-4)=3/(x+1)(x-2)+1
两边同乘(x+1)(x+2)(x-2)

[1/(x+2)(x+1)]+[x^2/(x-4)]-[3/(x-2)(x+1)]-1=0

(x^2-6x+8+x^5-4x^3+x^4-4x^2-x^2+2x+8-x^4+4x^2-x^3+4x+4x^3-16x+4x^2-16