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Ta có: \(x=2011\Rightarrow x+1=2012\)
Khi đó, ta có:
\(H\left(x\right)=x^4-\left(x+1\right).x^3+\left(x+1\right).x^2-\left(x+1\right).x+2012\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+2012\)
\(\Rightarrow H\left(2011\right)=-2011+2012=1\).
Vậy \(H\left(2011\right)=1\)
Cách 2:
\(H\left(x\right)=x^4-2012x^3+2012x^2-2012x+2012\)
\(=x^4-2011x^3-x^3+2011x^2+x^2-2011x-x+2011+1\)
\(=x^3\left(x-2011\right)-x^2\left(x-2011\right)+x\left(x-2011\right)-\left(x-2011\right)+1\)
\(=\left(x^3-x^2+x-1\right)\left(x-2011\right)+1\)
\(\Rightarrow H\left(2011\right)=1\)
Vậy...
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Ta có: x=2011 \(\Rightarrow\)x+1=2012
\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)
=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)
= \(x-1=2011-1=2010\)
=
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Ta có: \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=1+\left(1+\frac{2012}{2}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2014}{2014}+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(\Rightarrow x=2014\)
Lưu ý: số 2013 ở dòng T2 được tách ra làm 2013 số 1
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x=2012
nên x+1=2013
\(f\left(x\right)=x^{2013}-x^{2012}\left(x+1\right)+x^{2011}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}-...-x^3-x^2+x^2+x-1\)
=x-1
=2012-1=2011
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`(x+3)^2014 = (x+3)^2012`
`(x+3)^2014 -(x+3)^2012 =0`
`(x+3)^2012 [(x+3)^2 -1]=0`
TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`
TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}`
`=> [(x=-2),(x=-4):}`
`(x+3)^2014 = (x+3)^2012`
`=> (x+3)^2014 - (x+3)^2012 = 0`
`=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0`
`=> (x+3)^2012 * [ (x+3)^2 - 1] =0`
`=>`\(\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
Vậy, `x = {-3; -2; -4}.`
x^4-2012(x^3-x^2+x-1)
mà 2012=x
suy ra h(2012)=x^4-x.x^3+x.x^2-x.x+2012
=x^4-x^4+x^3-x^2+x
=x^3-x^2+x
=2012(2012^2-2012+1)
=2012(2012.2011+1)
=2012^2.2011+2012