已知方程ax2+bx+c=0(a≠0)的兩根為x1,x2.s1=x12005+x22005,s2=x12004+x22004,s3=x12003+x22003.求as1+bs2+cs3的值.
【答案】分析:已知方程ax2+bx+c=0(a≠0)的兩根為x1,x2,可將根代入方程得ax12+bx1+c=0,ax22+bx2+c=0,再將所得代入下式得as1+bs2+cs3=a(x12005+x22005)+b(x12004+x22004)+c(x12003+x22003)=(ax12005+bx12004+cx12003)+(ax22005+bx22004+cx22003)=x12003(ax12+bx1+c)+x22003(ax22+bx2+c)=x12003×0+x22003×0=0.
解答:解:∵x1,x2是方程ax2+bx+c=0(a≠0)的兩根.
∴ax12+bx1+c=0,ax22+bx2+c=0
又∵s1=x12005+x22005,s2=x12004+x22004,s3=x12003+x22003
∴as1+bs2+cs3=a(x12005+x22005)+b(x12004+x22004)+c(x12003+x22003)
=ax12005+ax22005+bx12004+bx22004+cx12003+cx22003
=(ax12005+bx12004+cx12003)+(ax22005+bx22004+cx22003)
=x12003(ax12+bx1+c)+x22003(ax22+bx2+c)
=x12003×0+x22003×0
=0
點評:本題不僅考查代數(shù)式的求值問題,還重點考查了一元二次方程的解得問題,再結(jié)合因式分解的方法,化簡代數(shù)式,本題有點難度,要認(rèn)真分析,聯(lián)系已知,要引起注意.