Đáp án:
$d)\sqrt{2}\\ e)0\\ f)2\sqrt{2}$
Giải thích các bước giải:
$d)\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\\ =\dfrac{\sqrt{2}\left(\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{16-6\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\\ =\dfrac{\sqrt{9-2.3.\sqrt{7}+7}+\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\\ =\dfrac{\sqrt{(3-\sqrt{7})^2}+\sqrt{(\sqrt{7}-1)^2}}{\sqrt{2}}\\ =\dfrac{3-\sqrt{7}+\sqrt{7}-1}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}\\ =\sqrt{2}\\ e)\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\\ =\dfrac{\sqrt{2}\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\\ =\dfrac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+2}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}+2}{\sqrt{2}}\\ =\dfrac{\sqrt{7}-1-(\sqrt{7}+1)+2}{\sqrt{2}}\\ =\dfrac{0}{\sqrt{2}}\\ =0\\ f)\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\\ =\dfrac{\sqrt{2}\left(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\\ =\dfrac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{11}+1)^2}-\sqrt{(\sqrt{11}-1)^2}+6}{\sqrt{2}}\\ =\dfrac{\sqrt{11}+1-(\sqrt{11}-1)+6}{\sqrt{2}}\\ =\dfrac{4}{\sqrt{2}}\\ =2\sqrt{2}$
