a) \(A=2x^2-3x-7=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{65}{8}=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{65}{8}\)
Do \(\left(x-\dfrac{3}{4}\right)^2\ge0\Rightarrow A\ge-\dfrac{65}{8}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
Vậy Min A = \(-\dfrac{65}{8}\) đạt được khi \(x=\dfrac{3}{4}\)
b) \(B=\left(x-2\right)^2+\left(x-3\right)^2=x^2-4x+4+x^2-6x+9=2x^2-10x+13\)
\(\Rightarrow B=2\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{1}{2}=2\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{2}\)
Do \(\left(x-\dfrac{5}{2}\right)^2\ge0\Rightarrow B\ge\dfrac{1}{2}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy Min B = \(\dfrac{1}{2}\) đạt được khi \(x=\dfrac{5}{2}\)
c) \(C=\left(x-2\right)^2+\left(y-3\right)^2\)
Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\Rightarrow C\ge0\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy min C = 0 đạt được khi \(x=2;y=3\)
