\(A=\dfrac{2x+7}{\sqrt{x}-1}=\dfrac{2x-2+9}{\sqrt{x}-1}=2\left(\sqrt{x}+1\right)+\dfrac{9}{\sqrt{x}-1}=2\left(\sqrt{x}-1\right)+\dfrac{9}{\sqrt{x}-1}+4\left(x>1\right)\)Ap dung BDT Cauhy cho cac so duong , ta co :
\(2\left(\sqrt{x}-1\right)+\dfrac{9}{\sqrt{x}-1}\text{≥}2\sqrt{2\left(\sqrt{x}-1\right).\dfrac{9}{\sqrt{x}-1}}=6\sqrt{2}\)
⇔ \(2\left(\sqrt{x}-1\right)+\dfrac{9}{\sqrt{x}-1}+4\text{≥ }6\sqrt{2}+4\)
⇒ \(A_{MIN}=6\sqrt{2}+4."="\text{⇔}\) \(2\left(\sqrt{x}-1\right)=\dfrac{9}{\sqrt{x}-1}\) ( tu giai ra nhe ) .
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