f(x)=(a+1)^2+2bx+(c-3)且 f(1)=f(3)=f(5)=4 a+b+c=?
回答 (2)
2015-10-31 4:57 pm
f(1)=(a+1)+2b+(c-3)=4
f(3)=9(a+1)+6b+(c-3)=4
f(5)=25(a+1)+10b+(c-3)=4
==>a=-1,b=0,c=7
a+b+c=-1+0+7=6
f(3)=9(a+1)+6b+(c-3)=4
f(5)=25(a+1)+10b+(c-3)=4
==>a=-1,b=0,c=7
a+b+c=-1+0+7=6
2015-11-01 12:04 am
f(x)=(a+1)x^2+2bx+(c－3)且 f(1)=f(3)=f(5)=4 a+b+c=?
Sol
degf(x)<=2
f(1)=f(3)=f(5)=4
f(x)=4
(a+1)x^2+2bx+(c－3)=4
a+1=0，2b=0，c－3=4
a=－1，b=0，c=7
a+b+c=6
Sol
degf(x)<=2
f(1)=f(3)=f(5)=4
f(x)=4
(a+1)x^2+2bx+(c－3)=4
a+1=0，2b=0，c－3=4
a=－1，b=0，c=7
a+b+c=6
收錄日期: 2021-04-30 20:16:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151031075314AAucOnq