Question

求答案，已知X是整数且X+3分之2+3-X分之2+X的平方-9分之2X+18位整数，求所有符合条件的X的值的和

Answer 1

2/(x+3)+2/(3-x)+(2x+18)/(x^2-9)
=2/(x+3)-2/(x-3)+(2x+18)/(x^2-9)
=2(x-3)/(x^2-9)-2(x+3)/(x^2-9)+(2x+18)/(x^2-9)
=(2x-6-2x-6+2x+18)/(x^2-9)
=(2x+6)/(x^2-9)
=2(x+3)/(x-3)(x+3)
=2/(x-3).
所以x-3只能取正负1，正负2 所以x=1、2、3、4

Answer 2

他们都不对，应该这样
解：题中(x+3)分之2+(3-x)分之2+(x²-9)分之(2x+18)也应为整数。
将上式化简后得： 2/(x+3)+2/(3-x)+(2x+18)/(x^2-9)
=2/(x+3)-2/(x-3)+(2x+18)/(x^2-9)
=2(x-3)/(x^2-9)-2(x+3)/(x^2-9)+(2x+18)/(x^2-9)
=(2x-6-2x-6+2x+18)/(x^2-9)
=(2x+6)/(x^2-9)
=2(x+3)/(x-3)(x+3)
=2/(x-3).
原式=2/(x-3)
符合条件的x的值有1、2、4、5
故所求和为：12

Answer 3

2/(x+3)+2/(3-x)+(2x+18)/(x^2-9)
=2/(x+3)-2/(x-3)+(2x+18)/(x^2-9)
=2(x-3)/(x^2-9)-2(x+3)/(x^2-9)+(2x+18)/(x^2-9)
=(2x-6-2x-6+2x+18)/(x^2-9)
=(2x+6)/(x^2-9)
=2(x+3)/(x-3)(x+3)
=2/(x-3).

Answer 4

看不懂你的题目，