因为(x+1) *(x-1)=x^2-1 所以方程可变形为
(x-2)(x+1) + (x+3)(x-1) + 2 / x^2-1=1
x^2-2x+x-2+x^2+3x-x-3+2 = x^2-1
x^2+x-2=0
x(x+1)=2
x=1
(x-2)/(x-1)+(x+3)/(x+1)+2/(x^2-1)=1
即(x-2)(x+1)/(x^2-1)+(x+3)(x-1)/(x^2-1)+2/(x^2-1)-(x^2-1)/(x^2-1)
=(x^2-x-2+x^2+2x-3+2-x^2+1)/(x^2-1)
=(x^2+x-2)/(x^2-1)
=(x+2)(x-1)/(x^2-1)
=(x+2)/(x+1)
=0
则x=-2或x=-1(舍去)
解:左面通分,得(x2+x-3)/(x2-1)=1,去分母,得x2+x-3=x2-1(x2-1≠0),所以x=2。