\(A=x^2-8x+17\)
\(=\left(x^2-8x+16\right)+1\)
\(=\left(x-4\right)^2+1\ge1\)
Dấu = xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy \(Min_A=1\Leftrightarrow x=4\)
\(B=x^2-x+1\)
\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_B=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(C=4x^2-12x+13\)
\(=\left(4x^2-12x+9\right)+4\)
\(=\left(2x-3\right)^2+4\ge4\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{3}{2}\)
Vậy \(Min_C=4\Leftrightarrow x=\frac{3}{2}\)
\(D=x^2-2x+y^2+4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(Min_D=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(E=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1-1\)
\(=\left(x^2+5x+5\right)^2-1\ge-1\)
\(\Rightarrow E_{min}=-1\) khi \(x^2+5x+5=0\Leftrightarrow x=\frac{-5\pm\sqrt{5}}{2}\)
