+) ta có : \(A=\sqrt{13+4\sqrt{10}}-\sqrt{13-4\sqrt{10}}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}-2\sqrt{2}+\sqrt{5}=2\sqrt{5}\) (sữa đề)
+) ta có : \(B=\sqrt{\dfrac{3-2\sqrt{2}}{17-12\sqrt{2}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(3-2\sqrt{2}\right)^2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{1}{\sqrt{2}-1}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{2}+1+2-\sqrt{3}=3-\sqrt{3}+\sqrt{2}\) (sữa đề )
+) đk : \(x\ne-3\)
ta có : \(C=\dfrac{\sqrt{x^2+6x+9}}{x+3}=\dfrac{\sqrt{\left(x+3\right)^2}}{x+3}=\dfrac{\left|x+3\right|}{x+3}\)
\(\left[{}\begin{matrix}C=1\left(x>-3\right)\\C=-1\left(x< -3\right)\end{matrix}\right.\)
+) \(m\ge\dfrac{5}{2}\)
ta có : \(D=\sqrt{2m+4+6\sqrt{2m-5}}-\sqrt{2m-5}\)
\(=\sqrt{\left(\sqrt{2m-5}+3\right)^2}-\sqrt{2m-5}=\left|\sqrt{2m-5}+3\right|-\sqrt{2m-5}\)
\(\Leftrightarrow\left[{}\begin{matrix}C=3\left(m\ge7\right)\\C=-3-2\sqrt{2m-5}\left(\dfrac{5}{2}\le m\le7\right)\end{matrix}\right.\)
