đại số 8

C

congchuaanhsang

Xét $F_{(f_{(x)}+x)}$=$(f_{(x)}+x)^2$+p($f_{(x)}$+x)+q
=$f_{(x)}^2$+2x$f_{(x)}$+$x^2$+p$f_{(x)}$+px+q
=$f_{(x)}$($f_{(x)}$+2x+p)+($x^2$+px+q)
=$f_{(x)}$($f_{(x)}$+2x+p+1)=$f_{(x)}$($x^2$+px+q+2x+p+1)
=$f_{(x)}$[$(x+1)^2$+p(x+1)+q]=$f_{(x)}$.$f_{(x+1)}$ với mọi x
Với x=2008 ta được $F_{(f_{(2008)}+2008)}$=$f_{(2008)}$.$f_{(2009)}$
Vậy k=$f_{(2008)}$+2008
 
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