ĐK: \(x\ge0\)
\(A=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1}\)
\(A=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
Vì \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0\Rightarrow\dfrac{4}{\sqrt{x}+1}>0\)
Áp dụng BĐT Cô-si ta có:
\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right)\left(\dfrac{4}{\sqrt{x}+1}\right)}=2\sqrt{4}=4\)
\(\Rightarrow\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\ge4-2=2\)
Dấu "=" xảy ra \(\sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=2\\\sqrt{x}+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-3\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
Vậy min A = 2 \(\Leftrightarrow\) x = 1
ĐKXĐ: x>=0
\(A=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}-2\)
=>\(A>=2\cdot\sqrt{4}-2=2\)
Dấu '=' xảy ra khi \(\sqrt{x}+1=\sqrt{4}=2\)
=>x=1
