Question

tìm A min

A= \(\dfrac{x+3}{\sqrt{x}}\)

Answer 1

T có A=\(\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}\)

Áp dụng Bđt cô si

Taco\(A=\sqrt{x}+\dfrac{3}{\sqrt{x}}\ge2\sqrt{\dfrac{3}{\sqrt{x}}.\sqrt{x}}=2\sqrt{3}\)

Vậy\(Min_A=2\sqrt{3}\)

Dấu '=' xảy ra <=>x=0

Answer 2

Áp dụng BĐT cosi cho \(x>0\left(ĐKXĐ\right)\)

\(ĐK:x>0\\ A=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\cdot\dfrac{3}{\sqrt{x}}}=2\sqrt{3}\\ A_{min}=2\sqrt{3}\Leftrightarrow\sqrt{x}=\dfrac{3}{\sqrt{x}}\Leftrightarrow x^2=3\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)