Ta có \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\Leftrightarrow C=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{3-2\sqrt{2}+16}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\\ =\dfrac{19-2\sqrt{2}}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{2-\sqrt{2}}\\ =\dfrac{\left(19-2\sqrt{2}\right)\left(2+\sqrt{2}\right)}{2}=\dfrac{34+15\sqrt{2}}{2}\)
Ta có \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\Leftrightarrow C=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{3-2\sqrt{2}+16}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\\ =\dfrac{19-2\sqrt{2}}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{2-\sqrt{2}}\\ =\dfrac{\left(19-2\sqrt{2}\right)\left(2+\sqrt{2}\right)}{2}=\dfrac{34+15\sqrt{2}}{2}\)