Question

用换元法解方程

Answer 1

x^2-x-12/(x^2-x)=4
令a=x^2-x
a-12/a=4
a^2-12=4a
a^2-4a-12=0
(a-6)(a+2)=0
a=6或a=-2

x^2-x=6
x^2-x-6=0
(x-3)(x+2)=0
x=3或x=-2

x^2-x=-2
x^2-x+2=0
△=1^2-4*1*2
=-7<0
所以方程无解

经检验x=3或x=-2是方程的解

3x^2/(x+1)-(x+1)/x^2=2
令x^2/(x+1)=m
3m-1/m=2
3m^2-1=2m
3m^2-2m-1=0
(3m+1)(m-1)=0
m=-1/3或m=1

x^2/(x+1)=-1/3
3x^2=-x-1
3x^2+x+1=0
△=1^2-4*3*1
=-11<0
所以方程无解

x^2/(x+1)=1
x^2=x+1
x^2-x-1=0
x^2-x+1/4-1/4-1=0
(x-1/2)^2-5/4=0
(x-1/2-√5/2)(x-1/2+√5/2)=0
x=1/2+√5/2或x=1/2-√5/2
经检验x=1/2+√5/2或x=1/2-√5/2是方程的解

△=1^2-4*3*1
=-11<0
所以方程无解

Answer 2

1)设t=x^2-x, 则方程化为：t-12/t=4, 即：t^2-4t-12=0,
(t-6)(t+2)=0
t=6, 得x^2-x-6=0, 得：(x-3)(x+2)=0, 得：x=3, -2
t=-2, 得x^2-x+2=0, 无实根
因此原方程的解为：x=3,-2
2)令t=x^2/(x+1), 则方程化为：3t-1/t=2, 即：3t^2-2t-1=0
(3t+1)(t-1)=0
t=-1/3, 得：x^2/(x+1)=-1/3, 得：3x^2+x+1=0, 无解
t=1， 得：x^2/(x+1)=1, 得：x^2-x-1=0, 得：x=(1+√5)/2, (1-√5)/2
因此原方程的解为：x=(1+√5)/2, (1-√5)/2

Answer 3

设t=x^2-x, 则方程化为：t-12/t=4, 即：t^2-4t-12=0,
(t-6)(t+2)=0
t=6, 得x^2-x-6=0, 得：(x-3)(x+2)=0, 得：x=3, -2
t=-2, 得x^2-x+2=0, 无实根
因此原方程的解为：x=3,-2
2)令t=x^2/(x+1), 则方程化为：3t-1/t=2, 即：3t^2-2t-1=0
(3t+1)(t-1)=0
t=-1/3, 得：x^2/(x+1)=-1/3, 得：3x^2+x+1=0, 无解
t=1， 得：x^2/(x+1)=1, 得：x^2-x-1=0, 得：x=(1+√5)/2, (1-√5)/2
解得：x=(1+√5)/2, (1-√5)/2