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a: Sửa đề: \(A=\sqrt{7-\sqrt{24}}+\sqrt{7+\sqrt{24}}\)
\(=\sqrt{6}-1+\sqrt{6}+1=2\sqrt{6}\)
b: \(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
Rút gọn:
\(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}.\left(\sqrt{6}-\sqrt{2}\right)}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].6}-\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].6}-\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}-1\right).6}-\sqrt{\left(3+\sqrt{3}-1\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(2+\sqrt{3}\right).6}-\sqrt{\left(2+\sqrt{3}\right).2}}{2}\)
= \(\dfrac{\sqrt{12+6\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\)
= \(\dfrac{3+\sqrt{3}-\left(1+\sqrt{3}\right)}{2}\)
= \(\dfrac{3+\sqrt{3}-1-\sqrt{3}}{2}\)
= \(\dfrac{2}{2}\)
= \(1\)
B=\(2\sqrt{18}-4\sqrt{32}+\sqrt{72}+3\sqrt{8}\)
\(=6\sqrt{2}-16\sqrt{2}+6\sqrt{2}+6\sqrt{2}\)
\(=2\sqrt{2}\)
\(B=2\sqrt{18}-4\sqrt{32}+\sqrt{72}+3\sqrt{8}\)
\(=6\sqrt{2}-16\sqrt{2}+6\sqrt{2}+6\sqrt{2}\)
\(=\sqrt{2}\left(6-16+6+6\right)\)
\(=2\sqrt{2}\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
= 5
\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)
\(=3\)