a.
Đặt \(\sqrt{x}+1=t\Rightarrow t\ge3\)
\(\sqrt{x}=t-1\)
\(\Rightarrow D=\dfrac{\left(t-1\right)^2-\left(t-1\right)+2}{t}=\dfrac{t^2-3t+4}{t}=t+\dfrac{4}{t}-3\)
\(D=\dfrac{4t}{9}+\dfrac{4}{t}+\dfrac{5t}{9}-3\ge2\sqrt{\dfrac{16t}{9t}}+\dfrac{5}{9}.3-3=\dfrac{4}{3}\)
\(D_{min}=\dfrac{4}{3}\) khi \(t=3\) hay \(x=4\)
b.
Đặt \(\sqrt{x}+2=t\Rightarrow t\ge4\)
\(\Rightarrow\sqrt{x}=t-2\)
\(M=\dfrac{\left(t-2\right)^2+8}{t}=\dfrac{t^2-4t+12}{t}=t+\dfrac{12}{t}-4\)
\(M=\dfrac{3t}{4}+\dfrac{12}{t}+\dfrac{1}{4}t-4\)
\(M\ge2\sqrt{\dfrac{36t}{4t}}+\dfrac{1}{4}.4-4=3\)
\(M_{min}=3\) khi \(t=4\) hay \(x=4\)
b.
\(M=\frac{x+8}{\sqrt{x}+2}=\frac{\sqrt{x}(\sqrt{x}+2)-2(\sqrt{x}+2)+12}{\sqrt{x}+2}=\sqrt{x}-2+\frac{12}{\sqrt{x}+2}\)
\(=\frac{3}{4}(\sqrt{x}+2)+\frac{12}{\sqrt{x}+2}+\frac{\sqrt{x}}{4}-\frac{7}{2}\)
\(\geq 2\sqrt{\frac{3}{4}.12}+\frac{\sqrt{4}}{4}-\frac{7}{2}=3\) (theo AM-GM)
Vậy $M_{\min}=3$ khi $x=4$