\(2x^2+2x+7=2x^2+2x+\frac{1}{2}+\frac{13}{2}\)
\(=2\left(x^2+x+\frac{1}{4}\right)+\frac{13}{2}=2.\left(x+\frac{1}{2}\right)^2+\frac{13}{2}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2.\left(x+\frac{1}{2}\right)^2+\frac{13}{2}\ge\frac{13}{2}\forall x\)
\(\Rightarrow2x^2+2x+7\ge\frac{13}{2}\forall x\)
hay biểu thức \(2x^2+2x+7\)luôn dương với mọi x ( đpcm )
2x2 + 2x + 7
= 2( x2 + x + 1/4 ) + 13/2
= 2( x + 1/2 )2 + 13/2 ≥ 13/2 > 0 ∀ x ( đpcm )
