c) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\) = \(\dfrac{\sqrt{12-8\sqrt{2}}}{\sqrt{2}}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
= \(\dfrac{\sqrt{\left(2\sqrt{2}-2\right)^2}}{\sqrt{2}}+\sqrt{\left(3\sqrt{2}-1\right)^2}\) = \(\dfrac{2\sqrt{2}-2}{\sqrt{2}}+3\sqrt{2}-1\)
\(\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{\sqrt{2}}+3\sqrt{2}-1\) = \(2-\sqrt{2}+3\sqrt{2}-1\) = \(2\sqrt{2}+1\)
d )Đặt A = \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(\Leftrightarrow A^2=\left(\sqrt{12-3\sqrt{7}}\right)^2-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}+\left(\sqrt{12+3\sqrt{7}}\right)^2\)
\(\Leftrightarrow A^2=12-3\sqrt{7}-2\sqrt{144-63}+12+3\sqrt{7}\)
\(\Leftrightarrow A^2=24-2\sqrt{81}\)
\(\Leftrightarrow A^2=24-18=6\)
=> A = \(\sqrt{6}\)
Vậy \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=\sqrt{6}\)
c ) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\) \(=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}2+4}+\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
= \(4-\sqrt{2}+3\sqrt{2}-1=3+2\sqrt{2}\)
