8: A=x(x+1)(x+2)(x+3)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1-1\)
\(=\left(x^2+3x+1\right)^2-1>=-1\forall x\)
Dấu '=' xảy ra khi \(x^2+3x+1=0\)
=>\(x^2+3x+\dfrac{9}{4}=\dfrac{5}{4}\)
=>\(\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)
=>\(x+\dfrac{3}{2}=\pm\dfrac{\sqrt{5}}{2}\)
=>\(x=-\dfrac{3}{2}\pm\dfrac{\sqrt{5}}{2}\)
9: \(A=\left(x-1\right)\left(x-2\right)\cdot\left(x-3\right)\left(x-4\right)+2008\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+2008\)
\(=\left(x^2-5x+5\right)^2-1+2008\)
\(=\left(x^2-5x+5\right)^2+2007>=2007\forall x\)
Dấu '=' xảy ra khi \(x^2-5x+5=0\)
=>\(x=\dfrac{5\pm\sqrt{5}}{2}\)
10: \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36>=-36\forall x\)
Dấu '=' xảy ra khi \(x^2+5x=0\)
=>x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
7. \(...\Leftrightarrow A=\dfrac{3x^2+3-2}{x^2+1}\)
\(\Leftrightarrow A=\dfrac{3\left(x^2+1\right)-2}{x^2+1}\)
\(\Leftrightarrow A=3-\dfrac{2x}{x^2+1}\)
Ta lại có \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\Rightarrow\dfrac{2x}{x^2+1}\le1\Leftrightarrow-\dfrac{2x}{x^2+1}\ge-1\)
\(\Rightarrow A=3-\dfrac{2x}{x^2+1}\ge3-1=2\)
Vậy \(GTNN\left(A\right)=2\left(tại.x=1\right)\)
