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發表 skywalker 於 星期日 一月 07, 2007 7:43 pm

#ed_op#DIV#ed_cl#f(x)=(x-1)(x-2)(x-3)q(x)+a(x-2)(x-3)+b(x-2)+10#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(3)=b+10=23=>b=13#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(1)=2a-3=1=>a=2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以餘式為2(x-2)(x-3)+13(x-2)+10#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=2x^2+3x-4#ed_op#/DIV#ed_cl#

發表 訪客 於 星期一 二月 20, 2006 12:54 pm

#ed_op#DIV#ed_cl##ed_op#SPAN id=convert147159#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#多項函數y=f(x)之圖形過(1,1),(2,10),(3,23),則f(x)除以(x-1)(x-2)(x-3)的餘式為?#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#f(x)=(x-1)(x-2)(x-3)*q(x)+ax^2+bx+c#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#f(1)=1=a+b+c#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#f(2)=10=4a+2b+c#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#f(3)=23=9a+3b+c#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=1#ed_cl#a=2,b=3,c=-4#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl#

[數學]f(x)除以(x-1)(x-2)(x-3)

發表 GFIF 於 星期一 二月 20, 2006 11:46 am

多項函數y=f(x)之圖形過(1,1),(2,10),(3,23),則f(x)除以(x-1)(x-2)(x-3)的餘式為?