一道初一分式题
a为整数,
(a2-4a+4)(a^3+2) (a+1)(a2-a+1)
且---------------- - -------------的值为正整数,求a
a^3-6a2+12a-8 a-2
人气:153 ℃ 时间:2020-01-30 19:41:53
解答
原式=(a-2)²(a³+2)/[(a-2)(a²+2a+4)-6a(a-2)]-(a+1)(a²-a+1)/(a-2)
=(a-2)²(a³+2)/[(a-2)(a²-4a+4)]-(a+1)(a²-a+1)/(a-2)
=(a-2)²(a³+2)/(a-2)³-(a+1)(a²-a+1)/(a-2)
=(a³+2)/(a-2)-(a³+1)/(a-2)
=(a³+2-a³-1)/(a-2)
=1/(a-2)
是正整数
所以a-2是1的约数
所以a-2=1
a=3
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