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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 $$\begin{cases}p + 2 = 0\\ 5 + q + 2p = m\\ 5q = n\\ 2q + 5p = 0\end{cases}\Rightarrow \begin{cases}p = -2\\ q = 5\\ m = 6\\ n = 25 \end{cases} \Rightarrow mn = 150.$$
2. 若 $a + b = 10$, $a^3 + b^3 = 100$, 则 $a^2 + b^2 = ?$
解答: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$= (a + b)\left[(a + b)^2 - 3ab\right]$$ $$\Rightarrow ab = 30$$ $$\Rightarrow a^2 + b^2 = (a + b)^2 - 2ab = 40.$$
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. $$\begin{cases}f(2) = 16 + 8m + 2n - 16 = 0\\ f(1) = 1+ m + n - 16 = 0 \end{cases}$$ $$\Rightarrow \begin{cases}m = -5\\ n = 20 \end{cases}\Rightarrow mn = -100.$$
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 $$(a + b + c)^2 - (a^2 + b^2 + c^2) = 0$$ $$\Rightarrow 1 - 3abc = 1 - (ab + bc + ca) = 1\Rightarrow abc = 0.$$
5. 若 $2a = 6b = 3c$, 且 $ab + bc + ca = 99$, 则 $2a^2 + 12b^2 + 9c^2 = ?$
解答: $$\begin{cases}a = 3b\\ c = 2b \end{cases} \Rightarrow 3b^2 + 2b^2 + 6b^2 = 99 \Rightarrow b^2 = 9.$$ 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: $$\begin{cases}x - y = 1 + m\\ y - z = 1 - m \end{cases} \Rightarrow x - z = 2$$ $$\Rightarrow x^2+y^2+z^2 - xy - yz - zx = {1\over2}\left[(x - y)^2 + (y - z)^2 + (z - x)^2\right]$$ $$= {1\over2}\left[(m+1)^2 + (1 - m)^2 + 4\right] = {1\over2}(2m^2 + 6)= m^2 + 3.$$