\(A=\dfrac{x^4+x}{x^2-x+1}+1-\dfrac{2x^2+3x+1}{x+1}\)
\(=\dfrac{x\left(x^3+1\right)}{x^2-x+1}+1-\dfrac{\left(2x+1\right)\left(x+1\right)}{x+1}\)
\(=x\left(x+1\right)+1-2x-1=x^2+x+1-2x-1=x^2-x\)
\(=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
\(...\Leftrightarrow A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}+1-\dfrac{2x^2+2x+x+1}{x+1}\)
\(\Leftrightarrow A=\dfrac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}+1-\dfrac{\left(x+1\right)\left(2x+1\right)}{x+1}\)
\(\Leftrightarrow A=x\left(x+1\right)+1-\left(2x+1\right)\)
\(\Leftrightarrow A=x^2-x\)
\(\Leftrightarrow A=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\)
\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy \(GTNN\left(A\right)=-\dfrac{1}{4}\left(tại.x=\dfrac{1}{2}\right)\)
