解分式方程哈1/(x²+11x-8)+1/(x²+2x-8)+1/(x²-13x-8)=0

记t=x^2+2x-8则方程化为：1/(t+9x)+1/t+1/(t-9x)=0去分母：t(t-9x)+(t+9x)(t-9x)+t(t+9x)=03t^2-81x^2=0t=±3√3x即x^2+(2±3√3)x-8=0解得x=[(2±3√3)+√(63±12√3)]/2经检验,这4个根都为原方程的根....