\(B=2x^2+10x-1\)
\(\Rightarrow B=2\left(x^2+5x\right)-1\)
\(\Rightarrow B=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{27}{2}\)
\(\Rightarrow B=\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)
Ta có : \(2\left(x+\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\)
Dấu "=" xảy rak hi và chỉ khi \(\left(x+\frac{5}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{5}{2}=0\)
\(\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(Min_B=\frac{-27}{2}\Leftrightarrow x=\frac{-5}{2}\)
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