1) \(A=x^2+8x+15=\left(x^2+8x+16\right)-1=\left(x+4\right)^2-1\ge-1\)
\(minA=-1\Leftrightarrow x=-4\)
2) \(B=7x-x^2-5=-\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{29}{4}=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{29}{4}\le\dfrac{29}{4}\)
\(maxB=\dfrac{29}{4}\Leftrightarrow x=\dfrac{7}{2}\)
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Ta có: \(A=x^2+8x+15\)
\(=x^2+8x+16-1\)
\(=\left(x+4\right)^2-1\ge-1\forall x\)
Dấu '=' xảy ra khi x=-4
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