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1. If $a ^2 + 2a + 5$ is a $a ^4 + ma^2 + n$, what is the value of $mn $?
Answer:
The method of undetermined coefficients is solved.
Make $a ^4 + ma^2 + N = (a^2 + 2a + 5) (a^2 + PA + q) $, then $$\begin{cases}p + 2 = 0\\ 5 + q + 2p = m\\ 5q = n\\ 2q + 5p = 0\end{ Cases}\rightarrow \begin{cases}p = -2\\ q = 5\\ m = 6\\ n = \end{cases} \rightarrow mn = 150.$$
2. If $a + b = 10$, $a ^3 + b^3 = 100$, then $a ^2 + b^2 =? $
Answer: $ $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.$$
3. If the polynomial $a ^4 + ma^3 + na-16$ contains the due-type $ (a-2) $ and $ (A-1) $, then $mn =?$
Answer:
The solution of the formula theorem. $$\begin{cases}f (2) = + 8m + 2n-16 = 0\\ F (1) = 1+ m + n-16 = 0 \end{cases}$$ $$\rightarrow \begin{cases}m = -5\\ n = \end{cases}\rightarrow MN = -100.$$
4. Known $a ^3 + b^3 + c^3 = a^2 + b^2+ c^2 = A+b+c = 1$, then $abc =? $
Answer:
by $ $a ^3 + b^3 + c^3-3abc = (A + B + C) (a^2 + b^2 + c^2-ab-bc-ca) $$ and $$ (A + B + c) ^2-(a^2 + b^2 + c^2) = 0$$ $$\ RightArrow 1-3ABC = 1-(ab + BC + CA) = 1\rightarrow ABC = 0.$$
5. If $2a = 6b = 3c$, and $ab + BC + CA = 99$, then $2a^2 + 12b^2 + 9c^2 =? $
Answer: $$\begin{cases}a = 3b\\ c = 2b \end{cases} \rightarrow 3b^2 + 2b^2 + 6b^2 = \rightarrow b^2 = 9.$ $ so $$2a^2 + 12b^2 + 9c^2 = 18b^2 + 12b^2 + 36b^2 = 594.$$
6. Set $x-y = 1+m$, $y-z = 1-m$, then $x ^2+Y^2+Z^2-XY-YZ-ZX =? $
Answer: $$\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.$$
2016 APE Tutoring Junior High School Math contest training Camp homework answer-8