書裡還有很多值得準備的題目，google books有的我就將網址列出來
沒有的我就將內容抄出來，想知道更多題目可以到圖書館去找這本書。
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Prove that the number \( \displaystyle c=\root 3 \of {\frac{1}{9}}+\root 3 \of {-\frac{2}{9}}+\root 3 \of {\frac{4}{9}} \) is a zero of \( F(x)=x^3+\root 3 \of 6 x^2-1 \).
第44頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA44#v=onepage&q&f=true
求滿足下列等式的數a：\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\frac{1}{\root 3 \of a}(1-\root 3 \of 2+\root 3 \of 4) \)
(1995日本奧林匹克預選賽)
若\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\root 3 \of a+\root 3 \of b+\root 3 \of c \)，則\( a+b+c= \)？
(92高中數學能力競賽，高中數學101 P25)
110.9.23補充
\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\root 3 \of a+\root 3 \of b+\root 3 \of c \)，其中\(a\)、\(b\)、\(c \in Q\)。求\(a+b+c=\)
。
(100中科實中，
https://math.pro/db/viewthread.php?tid=1107&page=3#pid3377)
112.8.21補充
已知實數\(x\)滿足拉馬努金等式\(\displaystyle \root 3 \of{\root 3\of 2-1}=\frac{1-\root 3\of 2+\root 3 \of 4}{x}\)，求實數\(x\)的值。(須以最簡單形式表示)
(111高中數學能力競賽 第一區(花蓮高中)口試試題，
https://math.pro/db/thread-3782-1-1.html)
當\(\root 3\of{\root 3\of{16}-2}=\root 3\of a+\root 3\of b+\root 3\of c\)，其中\(a,b,c \in Q\)。求\(a+b+c\)。
(109第1學期中山大學雙週一題)
110.2.25補充
若\(a\)是一個有理數且滿足\(\displaystyle \frac{1}{\root 3 \of 4+\root 3 \of 2+a}=\alpha \root 3 \of 4+\beta \root 3 \of 2+\gamma\)，其中\(\alpha,\beta,\gamma\)為有理數。試求\(\alpha,\beta,\gamma\)(用\(a\)表示)
(109北科附工，
https://math.pro/db/viewthread.php?tid=3326&page=2#pid21223)
111.1.10補充
有理化\( \displaystyle \frac{1}{\root 3 \of 2+\root 3 \of 3+\root 3 \of 5} \)
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For the expression
\( \displaystyle Q(n)=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdot \cdot \frac{2n-1}{2n} \) ( \( n \ge 2 \) )
we will prove the bounds \( \displaystyle \frac{1}{2 \sqrt{n}}<Q(n)<\frac{1}{\sqrt{2n+1}} \).
第103頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA103#v=onepage&q&f=true
\( \displaystyle a_n=\frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \)，試求\( \displaystyle \lim_{n \to \infty}a_n \)。
(97文華高中，連結已失效h ttp://forum.nta.org.tw/examservice/showthread.php?t=47781)
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Show that for arbitary \( x,y,z \in R \) we have
\( \displaystyle (\frac{x}{2}+\frac{y}{3}+\frac{z}{6})^2 \le \frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{6} \) (43)
Furthermore,determine when equality occures in (43).
第129頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA129#v=onepage&q&f=true
求\( \displaystyle (\frac{x}{2}+\frac{y}{3}+\frac{z}{6})^2=\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{6} \)的所有整數解。
(90高中數學能力競賽 中彰投區試題)
h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2002_Taiwan_High_Taichung_02.pdf 連結已失效
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Show that the inequality \( \sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x}\le 12 \) holds for all values \( x \in R \) for which the left-hand side is defined.
SOLUTION. By (48) with \( n=3 \), \( u_1=\sqrt{x+1} \),\( u_2=\sqrt{2x-3} \), and \( u_3=\sqrt{50-3x} \)
we have \( (\sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x})^2 \le 3(x+1+2x-3+50-3x)=144 \),
from which, upon taking the square root, we obtain the result.
求函數\( y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x} \)的最大和最小值。
(2009大陸高中數學競賽)
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Given an aribitrary prime p, solve the Diophantine equation \( x^4+4^x=p \).(第245頁)
SOLUTION. For any integer \( x<0 \), \( 4^x \) is not an integer, and neither is \( x^4+4^x \).
Thus for any prime p, the given equation has no negative solution.
For \( x=0 \) we have \( 0^4+4^0=1 \), which is not a prime, and for \( x=1 \) we have \( 1^4+4^1=5 \), a prime.
We will now show that for any integer \( x\ge 2 \), the number \( x^4+4^x \) is composite. If \( x=2k \) is even, where \( k \in N \), then
\( x^4+4^x=2^4 k^4+4^{2k}=16(k^4+4^{2(k-1)}) \),
which is a composite number. If \( x=2k+1 \)( \( k \in N \) ) is odd, then
\( x^4+4^x=x^4+4 \cdot 4^{2k}=[x^4+4x^2(2^k)^2+4(2^k)^4]-4x^2(2^k)^2 \)
\( =[x^2+2(2^k)^2]^2-(2x \cdot 2^k)^2 \)
\( =[x^2+2x \cdot 2^k+2(2^k)^2][x^2-2x \cdot 2^k+2(2^k)^2] \)
\( =[(x+2^k)^2+2^{2k}][(x-2^k)^2+2^{2k}] \),
which is again compostie, since \( (x±2^k)^2+2^{2k}\ge 2^{2k} \ge 2^2 >1 \).
To summarize, for \( p=5 \) the given equation has the unique solution \( x=1 \),while there are no solutions for any prime \( p \ne 5 \)
Prove that if \( n>1 \) then \( n^4+4^n \) is compostie.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328143
設\( n,a \)為自然數，試證：有無限多個a使得\( n^4+a \)必不為質數
(藍藍天上一朵雲 歷屆教甄考題整理2005版第75題)
Prove that there are infinitely many natural numbers a with the following property: the number \( z=n^4+a \) is not prime for any natural number n.
(1969IMO，
http://www.artofproblemsolving.c ... p/1969_IMO_Problems)
Compute \( \displaystyle \frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)} \).
(1987AIME)
101.1.10補充
100卓蘭實驗高中，
https://math.pro/db/thread-1165-1-1.html
高中數學101 P26
