\(C=x^2+2x+1\dfrac{1}{2}\)
\(C=x^2+2x+\dfrac{3}{2}\)
\(C=\left(x^2+2x+1\right)+\dfrac{1}{2}\)
\(C=\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu "=" xảy ra `<=>x+1=0`
`<=>x=-1`
Vậy `Min_C=1/2` khi `x=-1`
\(C=x^2+2x+\dfrac{3}{2}\\ =x^2+2x+1+\dfrac{1}{2}\\ =\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\)
Dấu bằng xảy ra khi
\(x+1=0\\ x=-1\)
Vậy \(Min_C=\dfrac{1}{2}khix=-1\)
C = x² + 2x + 3/2
= x² + 2x + 1 +1/2
= x² + x + x + 1 + 1/2
= (x² + x) + (x + 1) + 1/2
= x(x + 1) + (x + 1) + 1/2
= (x + 1)(x + 1) + 1/2
= (x + 1)² + 1/2
Do (x + 1)² >= 0
Suy ra (x + 1)² + 1/2 >= 1/2
Vậy GTNN của C là 1/2 khi x = -1
