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Bài 11. a) Ta có: $\sqrt{6+2\sqrt5}=\sqrt{5+2\sqrt5+1}=\sqrt{(\sqrt5+1)^2}=\sqrt5+1$ $\sqrt{6-2\sqrt5}=\sqrt{5-2\sqrt5+1}=\sqrt{(\sqrt5-1)^2}=\sqrt5-1$ Vậy $\sqrt{6+2\sqrt5}+\sqrt{6-2\sqrt5}=\sqrt5+1+\sqrt5-1=2\sqrt5$ b) Ta có: $\sqrt{3+2\sqrt2}=\sqrt{2+2\sqrt2+1}=\sqrt{(\sqrt2+1)^2}=\sqrt2+1$ $\sqrt{6-4\sqrt2}=\sqrt{4-4\sqrt2+2}=\sqrt{(2-\sqrt2)^2}=2-\sqrt2$ Vậy $\sqrt{3+2\sqrt2}+\sqrt{6-4\sqrt2}=\sqrt2+1+2-\sqrt2=3$ c) Ta có: $\sqrt{24+8\sqrt5}=\sqrt{16+8\sqrt5+8}=\sqrt{(4+\sqrt5)^2}=4+\sqrt5$ $\sqrt{9-4\sqrt5}=\sqrt{4-4\sqrt5+5}=\sqrt{(2-\sqrt5)^2}=\sqrt5-2$ Vậy $\sqrt{24+8\sqrt5}+\sqrt{9-4\sqrt5}=4+\sqrt5+\sqrt5-2=2+2\sqrt5$ d) Ta có: $\sqrt{41-12\sqrt5}=\sqrt{9-12\sqrt5+25}=\sqrt{(3-\sqrt5)^2}=3-\sqrt5$ $\sqrt{41+12\sqrt5}=\sqrt{25+12\sqrt5+9}=\sqrt{(5+3\sqrt5)^2}=5+3\sqrt5$ Vậy $\sqrt{41-12\sqrt5}-\sqrt{41+12\sqrt5}=3-\sqrt5-(5+3\sqrt5)=3-\sqrt5-5-3\sqrt5=-2-4\sqrt5$