a) \(B=3x^2-5x+7=3\left(x^2-2.\dfrac{5}{6}x+\dfrac{25}{36}\right)+\dfrac{59}{12}=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{59}{12}\)
Do \(\left(x-\dfrac{5}{6}\right)^2\ge0\Rightarrow B\ge\dfrac{59}{12}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)
Vậy \(B_{min}=\dfrac{59}{12}\) đạt được khi \(x=\dfrac{5}{6}\)
b) \(E=-x^2-4x-y^2+2y=-\left(x^2+4x+4\right)-\left(y^2-2x+1\right)+5=-\left(x+2\right)^2-\left(y-1\right)^2+5\)
Do \(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\Rightarrow E\le5\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)
Vậy \(E_{max}=5\) đạt được khi \(x=-2;y=1\)
c) \(I=\dfrac{15}{6}x-x^2-14=-\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}\right)-\dfrac{199}{16}=-\left(x-\dfrac{5}{4}\right)^2-\dfrac{199}{16}\)
Do \(\left(x-\dfrac{5}{4}\right)^2\ge0\Rightarrow I\le-\dfrac{199}{16}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{4}\)
Vậy \(I_{max}=-\dfrac{199}{16}\) đạt được khi \(x=\dfrac{5}{4}\)
d) \(M=8x+\dfrac{3}{4}x^2+1=\dfrac{3}{4}\left(x^2+2.\dfrac{16}{3}x+\dfrac{256}{9}\right)-\dfrac{61}{3}=\dfrac{3}{4}\left(x+\dfrac{16}{3}\right)^2-\dfrac{61}{3}\)
Do \(\left(x+\dfrac{16}{3}\right)^2\ge0\Rightarrow M\ge-\dfrac{61}{3}\)
Dấu = xảy ra \(\Leftrightarrow x+\dfrac{16}{3}=0\Leftrightarrow x=-\dfrac{16}{3}\)
Vậy \(M_{min}=-\dfrac{61}{3}\) đạt được khi \(x=-\dfrac{16}{3}\).
