\(A=x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3\ge3\\ A_{min}=3\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\\ B=\dfrac{x+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{5}{\sqrt{x}+1}\\ B=\sqrt{x}+1+\dfrac{5}{\sqrt{x}+1}-2\ge2\sqrt{\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{x}+1}}-2=2\sqrt{5}-2\\ B_{min}=2\sqrt{5}-2\Leftrightarrow\left(\sqrt{x}+1\right)^2=5\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1+\sqrt{5}\\\sqrt{x}=-1-\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6-2\sqrt{5}\\x=6+2\sqrt{5}\end{matrix}\right.\)
\(C=x+2-4\sqrt{x+2}+4-6=\left(\sqrt{x+2}-2\right)^2-6\ge-6\\ C_{min}=-6\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x=2\\ D=\sqrt{\left(x+1\right)^2}+\sqrt{\left(3-x\right)^2}\\ D=\left|x+1\right|+\left|3-x\right|=\left|x+1+3-x\right|=4\\ D_{min}=4\Leftrightarrow\left(x+1\right)\left(3-x\right)\ge0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)\le0\)
Dấu \("="\Leftrightarrow x+1;x+3\) khác dấu
