Question

Tìm GTNN của B = x^2 +5x+6

Answer 1

Ta có: \(B=x^2+5x+6\)

\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)

\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)

Vậy: \(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{5}{2}\)

Answer 2

\(B=x^2+5x+6=\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{25}{4}+6=\left(x+\dfrac{5}{2}\right)^2-\dfrac{1}{4}\)

Vì \(\left(x+\dfrac{5}{2}\right)^2\ge0\) nên \(B\ge-\dfrac{1}{4}\)

Vậy GTNN của B là \(-\dfrac{1}{4}\)

Dấu = xảy ra \(\text{⇔}x+\dfrac{5}{2}=0\text{⇔}x=-\dfrac{5}{2}\)