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6 tháng 6 2019
a) \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\sqrt{10}+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\sqrt{10}-1\)
b) \(\sqrt{21-6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}-\sqrt{3}\)
HH
4 tháng 12 2020
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
28 tháng 6 2019
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
9 tháng 7 2017
\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)
\(\Leftrightarrow10\)
\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-7\)
14 tháng 11 2019
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}+\frac{\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{11-2\sqrt{30}+11+2\sqrt{30}}{\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2}\)
\(=\frac{22}{1}=22\)
14 tháng 11 2019
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\sqrt{6}^2+\sqrt{5}^2}\)
\(=\sqrt{6}^2-2\sqrt{6}.\sqrt{5}+\sqrt{5}^2+\sqrt{6}^2+2\sqrt{6}.\sqrt{5}+\sqrt{5}^2\)
\(=6+5+6+5=22\)
Đặt A = \(\sqrt{5-\sqrt{21}}-\sqrt{6-\sqrt{32}}\)
\(\sqrt{2}A=\sqrt{10-2\sqrt{7.3}}-\sqrt{2}\sqrt{6-2\sqrt{4.2}}\)
\(=\sqrt{7}-\sqrt{3}-\sqrt{2}\left(\sqrt{4}-\sqrt{2}\right)\)
\(=\sqrt{7}-\sqrt{3}-2\sqrt{2}+2\)
Vậy \(A=\frac{\sqrt{7}-\sqrt{3}-2\sqrt{2}+2}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{6}-8+2\sqrt{2}}{2}\)
\(A=\sqrt{5-\sqrt{21}}-\sqrt{6-4\sqrt{2}}\)
nên \(A=\sqrt{\frac{7}{2}-2\sqrt{\frac{7}{2}.\frac{3}{2}}+\frac{3}{2}}-\sqrt{4-4\sqrt{2}+2}=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(A=\frac{\sqrt{14}-\sqrt{6}}{2}-2+\sqrt{2}\)