Question

数学题！急！高悬赏!!!!!!!!!

若X的平方-4=0，求（2X-6）/（X的平方-4X+4）÷（12-4X）/（X的平方+6X+9）×（X-2)/(X+3)的值！要有过程！

Answer 1

解：
[(2x-6)/(x^2-4x+4)]÷[(12-4x)/(x^2+6x+9)]×[(x-2)/(x+3)]
=[(2x-6)/(x^2-4x+4)]×[(x^2+6x+9)/(12-4x)]×[(x-2)/(x+3)]
=[(2x-6)/(x-2)^2]×[(x+3)^2/4(3-x)]×[(x-2)/(x+3)]
={2(x-3)[(x+3)^2](x-2)}/{[(x-2)^2]4(3-x)(x+3)}
=-(x+3)/[2(x-2)]
=-(x+3)(x+2)/[2(x-2)(x+3)]
=-(x+3)(x+2)/[2(x^2-4)]
因为：x^2-4=0
所以：所求之式的分母为0
因此，所求之式无意义！