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10 tháng 7 2019
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
10 tháng 7 2019
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
15 tháng 7 2019
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)
12 tháng 2 2022
\(a=\dfrac{4\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}}{2}\)
\(=2\sqrt{\sqrt{5}-\sqrt{5}+1}=2\)
\(P=\left(2^5-7\cdot2^2-3\right)^{81}+19=1+19=20\)
NT
1
27 tháng 4 2020
\(\left(3\sqrt{2}+\sqrt{6}\right).\sqrt{6-3\sqrt{3}}\) =\(\sqrt{2}\left(3+\sqrt{3}\right).\sqrt{6-3\sqrt{3}}\) =\(\left(3+\sqrt{3}\right).\sqrt{2.\left(6-3\sqrt{3}\right)}\) =\(\left(3+\sqrt{3}\right).\sqrt{12-6\sqrt{3}}\) =\(\left(3+\sqrt{3}\right).\sqrt{\left(3\right)^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}\) = \(\left(3+\sqrt{3}\right).\sqrt{\left(3-\sqrt{3}\right)^2}\) = \(\left(3+\sqrt{3}\right).|3-\sqrt{3}|\) =\(\left(3+\sqrt{3}\right).\left(3-\sqrt{3}\right)\) = \(9-3=6\)
Rút gọn hả bạn
\(=\frac{2\sqrt{2}\left(1-\sqrt{3}\right)}{3\sqrt{2-\sqrt{3}}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2}.\sqrt{2+\sqrt{3}}}{3.\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2\left(2+\sqrt{3}\right)}}{3.\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{4+2\sqrt{3}}}{3.\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\frac{2\left(1-\sqrt{3}\right)\sqrt{\left(1+\sqrt{3}\right)^2}}{3.\sqrt{4-3}}\)
\(=\frac{2\left(1-\sqrt{3}\right)|1+\sqrt{3}|}{3\sqrt{1}}\)
\(=\frac{2\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}{3}\)
\(=\frac{2\left(1^2-\left(\sqrt{3}\right)^2\right)}{3}\)
\(=\frac{2.\left(-2\right)}{3}=\frac{-4}{3}\)