Ta có :
\(2x^2+4x+15=2\left(x^2+2x+\dfrac{15}{2}\right)=2\left(x^2+2.x.1+1^2+\dfrac{13}{2}\right)=2\left[\left(x+1\right)^2+\dfrac{13}{2}\right]=2\left(x+1\right)^2+13\)
Với mọi x ta có :
\(2\left(x+1\right)^2\ge0\)
\(\Leftrightarrow2\left(x+1\right)^2+13\ge13\)
Dấu "=" xảy ra khi :
\(\left(x+1\right)^2=0\Leftrightarrow x=-1\)
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