\(2x^2+10x-1=\\ 2\left(x^2+5x-\dfrac{1}{2}\right)\\ =2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}-\dfrac{27}{4}\right)\\ =2\left(\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{4}\right)\\ =2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\)
vì \(2\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\)vậy Min \(2x^2+10x-1\) \(=-\dfrac{27}{2}\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow x+\dfrac{5}{2}=0\Rightarrow x=-\dfrac{5}{2}\)
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