\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+2008\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+2008\)
Đặt \(t=x^2-5x+4\) thì ta có:
\(t\left(t+2\right)+2008=t^2+2t+2008\)
\(=t^2+2t+1+2007\)
\(=\left(t+1\right)^2+2007\ge2007\)
Xảy ra khi \(t=-1\Rightarrow x^2-5x+4=-1\Rightarrow\)\(x=\dfrac{5\pm\sqrt{5}}{2}\)
A=( x - 1 ) ( x - 2 ) ( x - 3 ) (x - 4 ) + 2008
A=( x2-5x+4) (x2-5x+6) +2008
A=( x2-5x+5-1) ( x2-5x+5+1) +2008
A=( x2-5x+5)2-1+2008
A=( x2-5x+5)2 +2007 (1)
Mà ( x2-5x+5)2 \(\ge0\forall x\)(2)
Từ (1) và (2)\(\Rightarrow A\ge2007\)
Dấu "=" xảy ra ví dụ khi x=\(\dfrac{5+\sqrt{5}}{2}\)
Vậy Min A bằng 2007
