当x=_______时,代数式(x+3)/(x-2)与(x^2-4)/(3x^2-27)的值互为倒数

(x+3)/(x-2)与(x^2-4)/(3x^2-27)的值互为倒数
则(x+3)/(x-2)*(x^2-4)/(3x^2-27)=1，
(x+3)/(x-2)*(x-2)(x+2)/[3(x-3)(x+3)]=1
(x+2)/[3(x-3)]=1
x=11/2,x=-3

要使(X+3)/(X-2)与(X2-4)/(3X2-27)互为倒数
即(X+3)/(X-2) 与(X2-4)/(3X2-27)的积为1.
∴[(X+3)/(X-2)] *[(X2-4)/(3X2-27)]=1
∴[(X+3)/(X-2)] *[(X-2)(X+2)/3(X-3)(X+3)]=1
∴(X+2)/3(X-3)=1
∴X+2=3(X-3)
∴X=11/2

(x+3)/(x-2)*(x^2-4)/(3x^2-27)=1
(x+3)(x+2)=3x^2-27
x^2+5x+6=3x^2-27
2x^2-5x-33=0
(2x-11)(x+3)=0
x=11/2,x=-3

当x=__11/2_____时,代数式(x+3)/(x-2)与(x^2-4)/(3x^2-27)的值互为倒数
(x+3)/(x-2)*(x^2-4)/(3x^2-27)=1
(x+3)/(x-2)*(x-2)(x+2)/[3(x-3)(x+3)]=1
(x+2)/[3(x-3)]=1
x+2=3(x-3)
3x-9=x+2
3x-x=2+9
2x=11
x=11/2
经检验x=11/2符合要求