Bài 1 :
\(P=x^2+3x+7\)
\(=x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{19}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy GTNN của P là : \(\dfrac{19}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
\(Q=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-2x-5x+10\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-7x\right)^2-100\ge-100\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-7x=0\Leftrightarrow x\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy GTNN của Q là : \(-100\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Bài 2 : \(A=4x-x^2+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left[\left(x-2\right)^2-7\right]\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy GTLN của A là : \(7\Leftrightarrow x=2\)
\(B=x-x^2\)
\(=-\left(x^2-x\right)\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy GTLN của \(B\) là : \(\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
