Question

用换元法解方程: 1、2/(x-1)+(x^2-1)/4=0 2、(x^2)+1/(x^2)+2[x+(1/x)]+8=0

Answer 1

1、题目应是
2/(x-1)+4/(x^2-1)/4=0
2/(x-1)+4/(x-1)(x+1)=0
2/(x-1)=t
x-1=2/t
x+1=2/t
+2
(x+1)/2=(1+t)/t
2/(x+1)=t/(1+t)
则方程变为
t+t*t/(1+t)=0
t(1+t)+t^2=0
2t^2+t=0
t(2t+1)=0
t=0(舍去)
t=-1/2
2/(x-1)=-1/2
1-x=4
x=-3
经验证x=-3是方程的解
2、(x^2)+1/(x^2)+2[x+(1/x)]+8=0
(x^2)+2+1/(x^2)+2[x+(1/x)]+6=0
(x+1/x)^2+2[x+(1/x)]+6=0
设(x+1/x)=t
t^2+2t+6=0
判别式=2^2-4*1*6=-20<0
方程无解
题目有误