回復 1# Amis 的帖子
7(III)：
先利用勘根知此三根位在區間\(\left( -1,0 \right),\left( 0,1 \right),\left( 5,6 \right)\)
再由條件\(\left| \frac{\beta }{\alpha } \right|<1<\gamma \)與二分法, 推出\(\alpha \in \left( -1,0 \right),\beta \in \left( 0,1 \right),\gamma \in \left( 5,6 \right)\)
故\(\left| \frac{\alpha }{\gamma } \right|<1,\left| \frac{\beta }{\gamma } \right|<1\), 所求為
\(\frac{\beta }{\gamma }\cdot \frac{1}{1+\frac{\beta }{\alpha }}\cdot \frac{1}{1+\frac{\alpha }{\gamma }}\cdot \frac{1}{1+\frac{\beta }{\gamma }}=\frac{\alpha \beta \gamma }{\left( \alpha +\beta \right)\left( \beta +\gamma \right)\left( \gamma +\alpha \right)}=\frac{-1}{f\left( 6 \right)}=\frac{-1}{19}\)
8. 一定有更快的做法，我這作法很暴力，毫無美感冏，看看就好…
\(\frac{{{x}^{2}}}{\left( y+z \right)}+\frac{{{y}^{2}}}{\left( z+x \right)}+\frac{{{z}^{2}}}{\left( x+y \right)}=\frac{{{x}^{2}}\left( z+x \right)\left( x+y \right)+{{y}^{2}}\left( y+z \right)\left( x+y \right)+{{z}^{2}}\left( y+z \right)\left( z+x \right)}{\left( y+z \right)\left( z+x \right)\left( x+y \right)}=0\) 考慮將上式因式分解，利用輪換的觀念，發現有x+y+z之因式，故得到
\(\frac{\left( x+y+z \right)\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}+xyz \right)}{\left( y+z \right)\left( z+x \right)\left( x+y \right)}=0\Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}+xyz=0\)
(i) 所求為
\(\frac{x}{\left( y+z \right)}+\frac{y}{\left( z+x \right)}+\frac{z}{\left( x+y \right)}=\frac{x\left( z+x \right)\left( x+y \right)+y\left( y+z \right)\left( x+y \right)+z\left( y+z \right)\left( z+x \right)}{\left( y+z \right)\left( z+x \right)\left( x+y \right)}\)
\(=\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}+xyz+\left( y+z \right)\left( z+x \right)\left( x+y \right)}{\left( y+z \right)\left( z+x \right)\left( x+y \right)}=1\)
103.06.13 謝謝鋼琴老師指正，中間有計算錯誤，已修正
(ii) 所求為\(\frac{{{x}^{2}}}{yz}+\frac{{{y}^{2}}}{zx}+\frac{{{z}^{2}}}{xy}=\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{xyz}=-1\)
這題值得再想想…..
[ 本帖最後由 hua0127 於 2014-6-13 10:16 PM 編輯 ]
