(X^2+1)/(x+1) + 3(X+1)/(x^2+1)=4
設(shè) (X^2+1)/(x+1)=t
t+3/t-4=0
t^2-4t+3=0
t=1,t=3
t=1==> (X^2+1)/(x+1)=1 ==>x^2-x==0,x=0,x=1
,t=3
==> (X^2+1)/(x+1)=3 ==>x^2-3x-2=0 x=（3+√17)/2 ,x=（3-√17)/2
解為 x=0,或x=1,或x=（3+√17)/2 ,或x=（3-√17)/2謝謝。。。。還有個(gè)方程。。x平方+3x-(x平方+3x)分之20=810/（x+y）+3/（x-y）=-5 15/（x+y)-2/（x-y) =-11/(x+y)=t, 1/(x-y)=s10t+3s=-5(1)15t-2s=-1(2)(1)*2+(2)*3: 65t=-13 t=-1/5s=-1x+y=-5,x-y=-12x=-6,x=-3,y=-2解為：x=-3,y=-2x平方+3x-(x平方+3x)分之20=8x^2+3x-20/(x^2+3x)=8x^2+3x=tt-20/t=8t^2-8t-20=0t=10,t=-2t=10==> x^2+3x-10=0==>x1=-5,x2=2t=2==> x^2+3x+2=0 ==>x3=-1,x4=-2x1=-5,x2=2,x3=-1,x4=-2