1, \(A=3x^2+5x-1\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{1}{3}\right)\)
\(=3\left(x^2+\dfrac{5}{6}.x.2+\dfrac{25}{36}-\dfrac{37}{36}\right)\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{37}{12}\ge\dfrac{-37}{12}\)
Dấu " = " khi \(3\left(x+\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{-5}{6}\)
Vậy \(MIN_A=\dfrac{-37}{12}\) khi \(x=\dfrac{-5}{6}\)
2,3 tương tự
4, \(A=2x^2+7x\)
\(=2\left(x^2+\dfrac{7}{4}.x.2+\dfrac{49}{16}-\dfrac{49}{16}\right)\)
\(=2\left(x+\dfrac{7}{4}\right)^2-\dfrac{49}{8}\ge\dfrac{-49}{8}\)
Dấu " = " khi \(2\left(x+\dfrac{7}{4}\right)^2=0\Leftrightarrow x=\dfrac{-7}{4}\)
Vậy \(MIN_A=\dfrac{-49}{8}\) khi \(x=\dfrac{-7}{4}\)
5, 6 tương tự
7, \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " khi \(\left(x^2+5x\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(MIN_A=-36\) khi x = 0 hoặc x = -5
8, \(A=x^2-4x+y^2-8x+6\)
\(=x^2-4x+4+y^2-8x+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(MIN_A=-14\) khi x = 2 và y = 4
