\(A=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{x-1+9}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\)Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}\) ≥ \(2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}=2.3=6\)
⇔ \(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\text{≥}6-2=4\)
⇒ \(A_{Min}=4."="\text{⇔}x=4\left(TM\right)\)