Question

M= \(\sqrt{\dfrac{x}{2\sqrt{x}-2}}\) Tìm Min

Answer 1

\(M=\sqrt{\dfrac{x}{2\sqrt{x}-2}}=\sqrt{\dfrac{x-1+1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1}{2\left(\sqrt{x}-1\right)}}\)

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

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

Theo BĐT cô si ta có :

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}=2\)

\(\Rightarrow M\ge\sqrt{\dfrac{1}{2}\left(2+2\right)}=\sqrt{2}\)

Vậy GTNN của M là \(\sqrt{2}\) . Dấu bằng xảy ra khi \(x=4\)