Question

Chứng minh rằng \(x^2+5x-3\ge-\dfrac{37}{4}\)

Answer 1

\(x^2+5x-3=\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{37}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\)

(đpcm)

Answer 2

\(x^2+5x-3\)

\(=x^2+5x+\dfrac{25}{4}-\dfrac{37}{4}\)

\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\)

Dấu "=" khi \(x=-\dfrac{5}{2}\)