Question

先化简再求值3－x／x＋3＋x的平方＋6x＋9／x的平方－9,其中x＝根号3

Answer 1

3－x／x＋3＋x的平方＋6x＋9／x的平方－9,其中x＝根号3
=(3-x)/(x+3)+(x²+6x+9)/(x²-9)
=(3-x)/(x+3)+(x+3)²/(x-3)(x+3)
=-(x-3)²/(x+3)(x-3)+(x+3)²/(x-3)(x+3)
=(-x²+6x-9+x²+6x-9)/(x-3)(x+3)
=12x/(x-3)(x+3)
=12*√3/(√3-3)(√3+3)
=12√3/(3-9)
=12√3/(-6)
=-2√3

Answer 2

(3-x)/(x+3)+(x²+6x+9)/(x²-9)
=(3-x)/(x+3)+(x+3)²/(x-3)(x+3)
=-(x-3)²/(x+3)(x-3)+(x+3)/(x-3)
=(-x²+6x-9+x²+6x-9)/(x-3)(x+3)
=12x/（x²-9)（x=√3)
=-2√3