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\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
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. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
\(A=\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{5\sqrt{5}+5-5-\sqrt{5}}{\sqrt{5^2}-1}=\frac{5\sqrt{5}-\sqrt{5}}{5-1}=\frac{4\sqrt{5}}{4}=\sqrt{5}\)
\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)
\(\Leftrightarrow10\)
\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-7\)
\(\sqrt{21-8\sqrt{5}}\)\(-\sqrt{21-4\sqrt{5}}\)
\(=\sqrt{16-2.4\sqrt{5}+5}\)\(-\sqrt{20-2\sqrt{20}+1}\)
\(=\sqrt{\left(4-\sqrt{5}\right)^2}\)\(-\sqrt{\left(\sqrt{20}-1\right)}\)
\(=4-\sqrt{5}-\left(\sqrt{20}-1\right)\)
\(=4-\sqrt{5}-\sqrt{20}+1\)
\(=5-\sqrt{5}-2\sqrt{5}\)
\(=5-3\sqrt{5}\)
Có: \(\frac{P}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\right)\)
\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)
\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{3+\sqrt{5}}{2\sqrt{5}+\sqrt{5}+1}-\frac{3-\sqrt{5}}{2\sqrt{5}+\sqrt{5}-1}\)
\(=\frac{3+\sqrt{5}}{3\sqrt{5}+1}-\frac{3-\sqrt{5}}{3\sqrt{5}-1}\)
\(=\frac{\left(3+\sqrt{5}\right)\left(3\sqrt{5}-1\right)-\left(3-\sqrt{5}\right)\left(3\sqrt{5}+1\right)}{\left(3\sqrt{5}+1\right)\left(3\sqrt{5}-1\right)}\)
\(=\frac{9\sqrt{5}-3+15-\sqrt{5}-9\sqrt{5}-3+15+\sqrt{5}}{9\cdot5-1}\)
\(=\frac{24}{44}=\frac{6}{11}\)
=>P=\(\frac{6}{11}\cdot\sqrt{2}=\frac{6\sqrt{2}}{11}\)
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mink làm hơi tắt phần nào k hiểu hói mink nhé