Question

解分式方程1/(x+5)+1/(x+4)-3/(x+3)-3/(x+2)+1/(x+1)+1/x=0

Answer 1

1/(x+5)+1/(x+4)-3/(x+3)-3/(x+2)+1/(x+1)+1/x=0
[1/(x+5)+1/x]+[1/(x+4)+1/(x+1)]-3[1/(x+3)+1/(x+2)]=0
(2x+5)/[x(x+5)]+(2x+5)/[(x+1)(x+4)]-3(2x+5)/[(x+3)(x+2)]=0
(2x+5){1/[x(x+5)]+1/[(x+1)(x+4)]-3/[(x+3)(x+2)]}=0
(2x+5)=0或1/[x(x+5)]+1/[(x+1)(x+4)]-3/[(x+3)(x+2)]=0
(2x+5)=0时x=-5/2
1/[x(x+5)]+1/[(x+1)(x+4)]-3/[(x+3)(x+2)]=0时,即t=x(x+5)=x^2+5x(≥-25/4)
故1/[x(x+5)]+1/[(x+1)(x+4)]-3/[(x+3)(x+2)]=1/t+1/(t+4)-3/(t+6)=0
(t+4)(t+6)+t(t+6)-3t(t+4)=0(t≠0,t≠-4,t≠-6)
t^2-4t-24=0,(t-2)^2=28,t=2±2√7,
t=2+2√7,x^2+5x=2+2√7,x=-5/2±√(29/4+2√7)
t=2-2√7,x^2+5x=2-2√7,x=-5/2±√(29/4-2√7)
故,x1=-5/2,x2=-5/2+√(29/4+2√7),x3=-5/2-√(29/4+2√7),x4=-5/2+√(29/4-2√7),x5=-5/2-√(29/4-2√7)