\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+5\)
\(\Leftrightarrow A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+5\)
\(\Leftrightarrow A=\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)+5\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+5\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)\left[\left(x^2-5x+4\right)+2\right]+5\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)+5\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2x^2-10x+8+5\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2x^2-10x+13\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2x^2-10x+\dfrac{25}{2}+\dfrac{1}{2}\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+\left(2x^2-10x+\dfrac{25}{2}\right)+\dfrac{1}{2}\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{1}{2}\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{1}{2}\)
\(\Leftrightarrow A=\left(x^2-5x+4\right)^2+2\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{2}\)
Vậy GTNN của \(A=\dfrac{1}{2}\) khi \(\left\{{}\begin{matrix}x^2-5x+4=0\\x-\dfrac{5}{2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+4=0\left(loai\right)\\x=\dfrac{5}{2}\end{matrix}\right.\)
