已知X为整数,且2⼀(X+3) +2⼀(3-X ) +(2X+18)⼀(X^2—9)的植为整数,确定符合条件的X的直

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2025-05-15 15:58:14
推荐回答(3个)
回答(1):

2/(X+3) +2/(3-X ) +(2X+18)/(X^2—9)
=2/(X+3) -2/(X-3 ) +(2X+18)/(X^2—9)
=2X-6-2X-6 +2X+18/(X^2—9)
=2(X+3)/(X-3)(X+3)

因为X不等于-3
所以,原式=2/(X-3)

如果令式子为整数,则有
X-3=1 X=4,
X-3=-1 X=2,
X-3=2 X=5,
X-3=-2 X=1

回答(2):

2/(X+3) +2/(3-X ) +(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)

所以x可以是1,2,4,5

回答(3):

2/(X+3) +2/(3-X ) +(2X+18)/(X^2—9)
= 2(X-3)/(X+3)(X-3) - 2(X+3)/(X+3)(X-3) +(2X+18)/(X+3)(X-3)
= (2X-6-2X-6+2X+18)/(X+3)(X-3)
= (2X+6)/(X+3)(X-3)
= 2/(X-3)

该值为整数,所以,(X-3)可以为2,1,-1,-2
相应的,X=5,4,2,1。