Question

Tìm GTNN của biểu thức sau:

C=\(\dfrac{x^2+5x+7}{\left(x+3\right)^2}\)

Answer 1

C = \(\dfrac{x^2+5x+7}{\left(x+3\right)^2}\)

= \(\dfrac{x^2+6x+9-\left(x+3\right)+1}{\left(x+3\right)^2}\)

= \(\dfrac{\left(x+3\right)^2}{\left(x+3\right)^2}\) - \(\dfrac{x+3}{\left(x+3\right)^2}\) + \(\dfrac{1}{\left(x+3\right)^2}\)

= 1 - \(\dfrac{1}{x+3}\) + \(\dfrac{1}{\left(x+3\right)^2}\)

Đặt \(\dfrac{1}{x+3}\)= y, ta có:

1 - y + y²

= (y² - y + \(\dfrac{1}{4}\)) + \(\dfrac{3}{4}\)

= (y - \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\)

Dấu "=" xảy ra khi:

y - \(\dfrac{1}{2}\) = 0

\(\Leftrightarrow\) y = \(\dfrac{1}{2}\)

\(\Leftrightarrow\) \(\dfrac{1}{x+3}\) = \(\dfrac{1}{2}\)

\(\Leftrightarrow\) x + 3 = 2

\(\Leftrightarrow\) x = -1

Vậy GTNN của C = \(\dfrac{3}{4}\) tại x = -1