Question

换元法解方程⑴ 2（x^2+1）/（x+1）+6（x+1）/（x^2+1）=7 ⑵（x^4+2x^2+1）/(x^2)+（x^2+1）/x=2需步骤

Answer 1

换元法解方程⑴ 2（x^2+1）/（x+1）+6（x+1）/（x^2+1）=7
令（x^2+1）/（x+1）=t
2t+6/t=7
2t^2-7t+6=0
(2t-3)(t-2)=0
t=3/2或t=2
(1)t=3/2
（x^2+1）/（x+1）=3/2
2（x^2+1）=3（x+1）
2x^2+2=3x+3
2x^2-3x-1=0
x=(3±√3^2+4*2*1)/2*2=(3±√17)/4
(2)t=2
（x^2+1）/（x+1）=2
（x^2+1）=2（x+1）
x^2+1-2x-2=0
x^2-2x-1=0
x=(2±√2^2+4*1*1)/2*1=(1±√2)

⑵（x^4+2x^2+1）/(x^2)+（x^2+1）/x=2
(x^2+1)^2/(x^2)+（x^2+1）/x=2
(x^2+1/x)^2+（x^2+1）/x=2
令x^2+1/x=t
t^2+t=2
t^2+t-2=0
(t-1)(t+2)=0
t=1或t=-2
(1)t=1
x^2+1/x=1
x^2+1=x
x^2-x+1=0
判别式=1^2-4*1*1=-3<0
所以方程无解

(2)t=-2
x^2+1/x=-2
x^2+1=-2x
x^2+2x+1=0
(x+1)^2=0
所以x=-1