\(=-3\left(x^2+3x+4\right)\\ =-3\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{7}{4}\right)\\ =-3\left(x+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\le-\dfrac{21}{4}\)
Dấu \("="\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
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