Question

Tìm GTNN của: \(A=\dfrac{\left(x+4\right).\left(x+9\right)}{x}\) với x>0

Answer 1

\(A=x+13+\dfrac{36}{x}=\left(x+\dfrac{36}{x}\right)+13\ge2\sqrt{\dfrac{x.36}{x}}+13=12+13=25.\text{ Dấu }"="\text{ xảy ra khi: }x=\dfrac{36}{x}\text{ hay: }x=6\)

Answer 2

Ta có: \(A=\dfrac{x^2+13x+36}{x}=\dfrac{25x+x^2-12x+36}{x}\) \(=\dfrac{25x+\left(x-6\right)^2}{x}=25+\dfrac{\left(x-6\right)^2}{x}\ge25\)

Dấu bằng xảy ra \(\Leftrightarrow x=6\)

 Vậy \(Min_A=25\) khi \(x=6\)