a: \(=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=0\)
b: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\sqrt{7}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}+\sqrt{7}\)
\(=\sqrt{7}-\sqrt{2}\)
Đúng 1
Bình luận (0)
c) \(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}+2\sqrt{6}\)
\(=\sqrt{\left(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\right)^2}+2\sqrt{6}\)
\(=\sqrt{6,5+12-2\sqrt{\left(6,5+\sqrt{12}\right)\left(6,5-\sqrt{12}\right)+6,5-12}}+2\sqrt{6}\)
\(=\sqrt{13-2\sqrt{6,5^2-12}}+2\sqrt{6}\)
\(=\sqrt{13-2.\dfrac{11}{2}}+2\sqrt{6}\)
\(=\sqrt{2}+2\sqrt{6}\)
Đúng 1
Bình luận (0)
