ĐK: \(x\ge0\)
Ta có:
M = \(\frac{x+6\sqrt{x}+34}{\sqrt{x}+3}\)
=\(\frac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}\)
= \(\frac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}\)
=\(\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\)
Áp dụng BĐT Cauchy cho hai số không âm ta có:
\(\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\frac{25}{\sqrt{x}+3}}=2.5=10\)
Hay \(M\ge10\)
Dấu '=' xảy ra \(\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\)
\(\Leftrightarrow\sqrt{x}+3=5\)(vì \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\))
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(TM\right)\)
Vậy,...
Học giỏi toán nhé! 
ĐK \(x\ge0\)
\(M=\frac{x+16\sqrt{x}+64-10\sqrt{x}-30}{\sqrt{x}+3}\)
\(M=\frac{\left(\sqrt{x}+8\right)^2-10\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(M=\frac{\left(\sqrt{x}+8\right)^2}{\sqrt{x}+3}-10\)
ta có điều kiện \(x\ge0\) vậy \(M_{min}\) khi x=0
\(M_{min}=\frac{\left(\sqrt{0}+8\right)^2}{\sqrt{0}+3}-10=\frac{64}{3}-10=\frac{34}{3}\)
vậy \(M_{min}=\frac{34}{3}\) khi x=0
