《2025年智慧学习明天出版社八年级数学上册人教版》

22. (6分) 已知实数$a$,$b$,$c$,$m$,$n满足3m + n = \frac{b}{a}$,$mn = \frac{c}{a}$.
(1) 求证：$b^{2} - 12ac$为非负数；
(2) 若$a$,$b$,$c$均为奇数,$m$,$n$是否可以都为整数？说明你的理由.

答案：
(1)$\because 3m+n=\frac{b}{a},mn=\frac{c}{a},\therefore b=a(3m+n),c=amn$,则$b^{2}-12ac=[a(3m+n)]^{2}-12a^{2}mn=a^{2}(9m^{2}+6mn+n^{2})-12a^{2}mn=a^{2}(9m^{2}-6mn+n^{2})=a^{2}(3m-n)^{2}$,$\because a,m,n$是实数,$\therefore a^{2}(3m-n)^{2}\geq 0$.$\therefore b^{2}-12ac$为非负数.
(2)$m,n$不可能都为整数.理由如下:若$m,n$都为整数,其可能情况有:①当$m,n$都为奇数时,则$3m+n$必为偶数,又$\because 3m+n=\frac{b}{a},\therefore b=a(3m+n)$.$\because a$为奇数,$\therefore a(3m+n)$必为偶数,这与$b$为奇数矛盾;②当$m,n$为整数,且其中至少有一个为偶数时,则$mn$必为偶数,又$\because mn=\frac{c}{a},\therefore c=amn$.$\because a$为奇数,$\therefore amn$必为偶数,这与$c$为奇数矛盾.综上所述,$m,n$不可能都为整数.