ĐKXĐ : \(x\ge0\)
Ta có \(A=\dfrac{x+2\sqrt{x}+16}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2+15}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{15}{\sqrt{x}+1}\ge2.\sqrt{\left(\sqrt{x}+1\right).\dfrac{15}{\sqrt{x}+1}}=2.\sqrt{15}\)
Dấu = xảy ra <=> \(\left(\sqrt{x}+1\right)^2=15\Leftrightarrow x=\left(\sqrt{15}-1\right)^2\)




giúp mik với cả 2 bài nha





