已知\( x+y+z=2 \),\( x^2+y^2+z^2=3 \),\( x^3+x^3+z^3=4 \),試求\( x^4+x^4+z^4 \)。
(我的教甄準備之路 利用根與係數的關係解聯立方程式)
https://math.pro/db/viewthread.php?tid=661&page=1#pid1076
已知\( \cases{a+b=8 \cr ax+by=9 \cr ax^2+by^2=57 \cr ax^3+by^3=111} \),求\( ax^4+by^4 \)
補充一題
設c,d,x,y為實數,滿足\( \cases{ax+by=3 \cr ax^2+by^2=7 \cr ax^3+by^3=16 \cr ax^4+by^4=42} \),求\( cx^5+dy^5 \)之值
http://forum.nta.org.tw/examservice/showthread.php?t=47266
http://forum.nta.org.tw/examservice/showthread.php?t=26666
Find \( ax^5+by^5 \) if the real numbers \( a,b,x \)and \(y\) satisfy the equations \( \cases{ax+by=3 \cr ax^2+by^2=7 \cr ax^3+by^3=16 \cr ax^4+by^4=42} \)
(1990AIME,
http://www.artofproblemsolving.c ... 82&cid=45&year=1990
99鳳新高中,
https://math.pro/db/thread-974-1-1.html)
這裡補充用遞迴的方法
令\( A_n=ax^n+by^n \)
\( ax^{n+1}+by^{n+1}=(x+y)(ax^n+by^n)-xy(ax^{n-1}+by^{n-1}) \)
\( \matrix{111=(x+y)(57)-xy(9) \cr 57=(x+y)(9)-xy(8)} \)
得\( x+y=1 \),\( xy=-6 \),\( A_{n+1}=A_n+6A_{n-1} \)
\( ax^4+by^4=(ax^3+by^3)+6(ax^2+by^2)=453 \)
2009.9.27補充
http://www.yll.url.tw/viewtopic.php?t=23885
2010.3.27補充
98學年度第二學期中山大學雙週一題
http://www.math.nsysu.edu.tw/~problem/2010s/2Q.pdf
2010.3.29補充
第3,4題
http://www.shiner.idv.tw/teachers/viewtopic.php?p=3014
設實數c,d,x,y滿足\( cx+dy=3 \),\( cx^3+dy^3=16 \)且\( cx^2+dy^2=3 \),\( cx^4+dy^4=16 \)試求:\( cx^5+dy^5 \)之值。
(94高中數學能力競賽 高屏區筆試一)
2010.6.27補充
The sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers?
http://www.artofproblemsolving.c ... .php?f=150&t=354265
101.6.28補充
若\( a,b,x,y \in R \),\( \displaystyle \cases{a+b=4 \cr ax+by=13 \cr ax^2+by^2=41 \cr ax^3+by^3=127} \),求\( ax^4+by^4 \)
(101中正高中二招,
https://math.pro/db/thread-1446-1-1.html)
103.5.15補充
若\( \displaystyle \cases{a+b=1 \cr ax+by=-1 \cr ax^2+by^2=-5 \cr ax^3+by^3=-13} \),求\( ax^5+by^5 \)之值為
。
(103彰化高中,
https://math.pro/db/thread-1890-1-1.html)
104.5.2補充
已知實數\( x,y,a,b \)滿足\( \cases{ax+by=1 \cr ax^2+by^2=2 \cr ax^3+by^3=8 \cr ax^5+by^5=100} \),則\( ax^4+by^4= \)
。
(104鳳山高中,
https://math.pro/db/thread-2244-1-1.html)