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Lộn đề
\(A=x^{2009}-2008x^{2008}-2008x^{2007}-...-2008x+1\)1
\(x=2009\)
\(\Rightarrow x-1=2008\left(1\right)\)
Thay (1) vào A ta được:
\(A=x^{2009}-2008x^{2008}-2008x^{2007}-...-2008x+1\)
\(A=x^{2009}-\left(x-1\right)x^{2008}-...-\left(x-1\right)x+1\)
\(A=x^{2009}-x^{2009}+x^{2008}-...-x^2-x+1\)
\(A=-x+1\)
\(A=-2009+1\)
\(A=-2008\)
x=2009x=2009
⇒x−1=2008(1)⇒x−1=2008(1)
Thay (1) vào A ta được:
A=x^2009−2008x^2008−2008x^2007−...−2008x+1
A=x^2009−(x−1)x^2008−...−(x−1)x+1
A=x^2009−x^2009+x^2008−...−x^2−x+1
A=−x+1
A=−2009+1
A=−2008
\(x=2009\Leftrightarrow x-1=2008\\ \Leftrightarrow A=x^x-\left(x-1\right)x^{x-1}-\left(x-1\right)x^{x-2}-...-\left(x-1\right)x+1\\ \Leftrightarrow A=x^x-x^x+x^{x-1}-x^{x-1}+x^{x-2}-...-x^2-x+1\\ \Leftrightarrow A=1-x=1-2009=-2008\)
a.\(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
Sửa đề:.\(x^4+2008x^2+2007x+2008\)
\(=x^4+x^2+1+2007x^2+2007x+2007\)
\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x+2008-x
=x(x^3-1)+2008(x^2+x+1)
=x(x-1)(x^2+x+1)+2008(x^2+x+1)
=(x^2+x+1)(x^2-x+2008)
x4+2008x2+2007x+2008
=(x4+x2+1)+(2007x2+2007x+2007)
=(x4+2x2+1-x2)+2007(x2+x+1)
=(x2+1)2-x2+2007(x2+x+1)
=(x2+1-x)(x2+1+x)+2007(x2+x+1)
=(x2+x+1)(x2+1-x+2007)=(x2+x+1)(x2-x+2008)
Đề sai rồi bạn
Nguyễn Lê Phước Thịnh CTV
chuyên toán