\(ĐK:\) \(x\ge0\)
\(P=\dfrac{x+8}{\sqrt{x}+1}\)
\(P=\dfrac{\left(x-1\right)+9}{\sqrt{x+1}}\)
\(P=\sqrt{x-1}+\dfrac{9}{\sqrt{x}+1}\)
\(P=\left(\sqrt{x+1}+\dfrac{9}{\sqrt{x+1}}\right)-2\ge2\)
\(P=\sqrt{\dfrac{\left(\sqrt{x}+1\right).9}{\sqrt{x+1}}-2=4}\) \((BĐT\) \(AM-GM)\)
Vậy...
\(P=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{\left(x-4\sqrt{x}+4\right)+4\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}+4\)Ta có \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\left(\sqrt{x}-2\right)^2\ge0\end{matrix}\right.\)⇒\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\ge0\Rightarrow\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}+4\ge4\Rightarrow P\ge4\)
Vậy MinP=4
Đẳng thức xảy ra khi \(\left(\sqrt{x}-2\right)^2=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(P=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{x-1+9}{\sqrt{x}+1}\)\(=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\)
Ta có: \(x\ge0\forall x\Rightarrow\sqrt{x}+1>0,\dfrac{9}{\sqrt{x}+1}>0\)
Áp dụng BĐT Cô-si ta có:
\(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}\)
\(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}\ge6\)
\(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\ge4\)
Dấu bằng xảy ra khi \(\sqrt{x}+1=\dfrac{9}{\sqrt{x}+1}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=-3\left(L\right)\\\sqrt{x}+1=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\left(TM\right)\)
Vậy MinA = 4 khi x = 4
