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一、計算題,第 5 題

已知 \(x,y,z\) 均為實數,且 \(\displaystyle \left\{ {\begin{array}{*{20}{c}}
   {{2^x} + {3^y} + {5^z} = 7}  \\
   {{2^{x - 1}} + {3^y} + {5^{z + 1}} = 11}  \\
\end{array}} \right.\),

若 \(t = {2^{x + 1}} + {3^y} + {5^{z - 1}}\),試求 \(t\) 的範圍.



解答:

令 \(\displaystyle a=2^x, b=3^y, c=5^z\),則

\(\displaystyle \left\{ {\begin{array}{*{20}{c}}
   {a + b + c = 7}  \\
   {\displaystyle \frac{a}{2} + b + 5c = 11}  \\
\end{array}} \right.\) 且 \(\displaystyle a,b,c>0\),


得此兩平面部分交線段的參數式 \(\displaystyle \left\{ {\begin{array}{*{20}{c}}
   {a = 0 + 8k}  \\
   {b = 6 - 9k}  \\
   {c = 1 + k}  \\
\end{array}} \right.\),


其中 \(\displaystyle a,b,c>0\Rightarrow 0<k<\frac{2}{3}\)


故,

\(\displaystyle t=2a+b+\frac{c}{5}=\frac{31+36k}{5}\)

\(\displaystyle \Rightarrow \frac{31}{5}<t<11.\)








第 8 題

設整數數多項式 \(A\left(x\right)\) 除以 \(x^2+1\),餘式為 \(px+q\),

若 \(f\left(A\left(x\right)\right)=pi+q\) 恆成立(其中 \(i\) 為虛數單位),

求 \(\displaystyle \frac{{f\left( {{x^{10}} + x + 1} \right)}}{{f\left( {{x^5} + x + 1} \right)}}\) 的值?


解答:

依題意,

因為 \(\displaystyle x^5+x+1\) 除以 \(\displaystyle x^2+1\) 的餘式為 \(2x+1\),

所以 \(\displaystyle f(x^5+x+1)=2i+1.\)

因為 \(\displaystyle x^{10}+x+1\) 除以 \(\displaystyle x^2+1\) 的餘式為 \(x\),

所以 \(\displaystyle f(x^{10}+x+1)=i.\)

故,

\(\displaystyle \frac{{f\left( {{x^{10}} + x + 1} \right)}}{{f\left( {{x^5} + x + 1} \right)}} = \frac{{i }}{2i+1} =\frac{2+i}{5}.\)

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