\(A=\dfrac{x+3}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}-1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}+2\)
\(\Rightarrow A\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{4}{\left(\sqrt{x}-1\right)}}+2=2.2+2=6\)
\(\Rightarrow A_{min}=6\) khi \(\left(\sqrt{x}-1\right)^2=4\Rightarrow x=9\)
Đặt A=\(\dfrac{x+3}{\sqrt{x}-1}\Leftrightarrow A\sqrt{x}-A=x+3\Leftrightarrow x-A\sqrt{x}+A+3=0\)
Để phương trình có nghiệm thì △=\(\left(-A\right)^2-4\left(A+3\right)\ge0\Leftrightarrow A^2-4A-12\ge0\Leftrightarrow A^2+2A-6A-12\ge0\Leftrightarrow A\left(A+2\right)-6\left(A+2\right)\ge0\Leftrightarrow\left(A+2\right)\left(A-6\right)\ge0\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}A+2\ge0\\A-6\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}A+2\le0\\A-6\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}A\ge-2\\A\ge6\end{matrix}\right.\\\left\{{}\begin{matrix}A\le-2\\A\le6\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}A\ge6\\A\le-2\end{matrix}\right.\)
Vậy GTNN của A là 6
Dấu '=' xảy ra khi \(\dfrac{x+3}{\sqrt{x}-1}=6\Leftrightarrow x+3=6\sqrt{x}-6\Leftrightarrow x-6\sqrt{x}+9=0\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\Leftrightarrow\sqrt{x}-3=0\Leftrightarrow x=9\)
