Sửa đề: \(\sqrt{7-2\sqrt{10}}-\sqrt{14+4\sqrt{10}}-2\sqrt{22-4\sqrt{10}}+3\cdot\sqrt{13-4\sqrt{10}}\)
Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{14+4\sqrt{10}}-2\sqrt{22-4\sqrt{10}}+3\cdot\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt5\cdot\sqrt2+2}-\sqrt{10+2\cdot\sqrt{10}\cdot2+4}-2\cdot\sqrt{20-2\cdot2\sqrt5\cdot\sqrt2+2}+3\cdot\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}\)
\(=\sqrt{\left(\sqrt5-\sqrt2\right)^2}-\sqrt{\left(\sqrt{10}+2\right)^2}-2\cdot\sqrt{\left(2\sqrt5-\sqrt2\right)^2}+3\cdot\sqrt{\left(2\sqrt2-\sqrt5\right)^2}\)
\(=\sqrt5-\sqrt2-\sqrt{10}-2-2\left(2\sqrt5-\sqrt2\right)+3\left(2\sqrt2-\sqrt5\right)\)
\(=\sqrt5-\sqrt2-\sqrt{10}-2-4\sqrt5+2\sqrt2+6\sqrt2-3\sqrt5=-6\sqrt5+7\sqrt2-\sqrt{10}\) -2
