giải phương trình:
Nếu \(x\ge1\)phương trình trở thành : \(x^2-3x+2=x-1\Leftrightarrow x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}TM}\)Nếu \(x< 1\)\(\Rightarrow x^2-3x+2=1-x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1L\)VẬY NGHIỆM PHƯƠNG TRÌNH LÀ : x=1 hoặc x=3
\(x^4+2008x^2+2007x+2008\)
\(=x\left[x\left(x^2+2008\right)+2007\right]+2008\)
\(=\left[\left(x-1\right)x+2008\right]\left(x^2+x+1\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
~(‾▿‾~)
\(x^4+2008x^2+2007x+2008\)
\(=x^4+2007x^2+x^2+2007x+2007+1\)
\(=\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1+2007\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
x^4 +2008.x^2+2007.x+2008= x^4 +2008x^2+2008x-x+2008
= 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)
