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1. 若 $a^2 + 2a + 5$ 是 $a^4 + ma^2 + n$ 的一个因式, 那么 $mn$ 的值是多少?
解答:
待定系数法求解.
令 $a^4 + ma^2 + n = (a^2 + 2a + 5)(a^2 + pa + q)$, 则 $$egin{cases}p + 2 = 0\ 5 + q + 2p = m\ 5q = n\ 2q + 5p = 0end{cases}Rightarrow egin{cases}p = -2\ q = 5\ m = 6\ n = 25 end{cases} Rightarrow mn = 150.$$
2. 若 $a + b = 10$, $a^3 + b^3 = 100$, 则 $a^2 + b^2 = ?$
解答: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$= (a + b)left[(a + b)^2 - 3ab ight]$$ $$Rightarrow ab = 30$$ $$Rightarrow a^2 + b^2 = (a + b)^2 - 2ab = 40.$$
3. 若多项式 $a^4 + ma^3 + na - 16$ 含有因式 $(a-2)$ 和 $(a-1)$, 则 $mn=?$
解答:
因式定理求解. $$egin{cases}f(2) = 16 + 8m + 2n - 16 = 0\ f(1) = 1+ m + n - 16 = 0 end{cases}$$ $$Rightarrow egin{cases}m = -5\ n = 20 end{cases}Rightarrow mn = -100.$$
4. 已知 $a^3 + b^3 + c^3 = a^2 + b^2+ c^2 = a+b+c = 1$, 则 $abc = ?$
解答:
由 $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ 及 $$(a + b + c)^2 - (a^2 + b^2 + c^2) = 0$$ $$Rightarrow 1 - 3abc = 1 - (ab + bc + ca) = 1Rightarrow abc = 0.$$
5. 若 $2a = 6b = 3c$, 且 $ab + bc + ca = 99$, 则 $2a^2 + 12b^2 + 9c^2 = ?$
解答: $$egin{cases}a = 3b\ c = 2b end{cases} Rightarrow 3b^2 + 2b^2 + 6b^2 = 99 Rightarrow b^2 = 9.$$ 因此 $$2a^2 + 12b^2 + 9c^2 = 18b^2 + 12b^2 + 36b^2 = 594.$$
6. 设 $x - y = 1+m$, $y-z = 1-m$, 则 $x^2+y^2+z^2 - xy - yz - zx = ?$
解答: $$egin{cases}x - y = 1 + m\ y - z = 1 - m end{cases} Rightarrow x - z = 2$$ $$Rightarrow x^2+y^2+z^2 - xy - yz - zx = {1over2}left[(x - y)^2 + (y - z)^2 + (z - x)^2 ight]$$ $$= {1over2}left[(m+1)^2 + (1 - m)^2 + 4 ight] = {1over2}(2m^2 + 6)= m^2 + 3.$$