Question

(x^2+2x-3)/(x^2-9)*(x^2-5x+6)/(3x^2-x-2)？

Answer 1

(x^2+2x-3)/(x^2-9)*(x^2-5x+6)/(3x^2-x-2)
什么题目?简化?
(x^2+2x-3)/(x^2-9)*(x^2-5x+6)/(3x^2-x-2)
=（x+3）（x-1）/[（x+3）（x-3）]*（x-2）（x-3）/[（3x+2）（x-1）]
=（x-2）/（3x+2）,9,(x^2+2x-3)=(X+3)(X-1)
(x^2-9)=(X+3)(X-3)
(x^2-5x+6)=(X-3)(X-2)
(3x^2-x-2)=(X-1)(3X+2)
所以原式等于:(X+3)(X-1)/(X+3)(X-3)*(X-3)(X-2)/(X-1)(3X+2)=(X-2)/(3X+2),1,(x^2+2x-3)/(x^2-9)*(x^2-5x+6)/(3x^2-x-2)
快,好的话分加倍!
/前是分子,/后是分母
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