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\(\sqrt{\frac{9}{\sqrt{14+4\sqrt{6}}}}-\sqrt{\frac{9}{\sqrt{14-4\sqrt{6}}}}\)
\(=\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2+2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}-\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2-2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}\)
\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)
\(=\frac{3}{\sqrt{\sqrt{12}+\sqrt{2}}}-\frac{3}{\sqrt{\sqrt{12}-\sqrt{2}}}=\frac{3\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)-3\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)}{\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)}\)
\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)}\)
\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{12-2}}\)
\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{10}}\)
\(=\frac{3\left(\sqrt{2\sqrt{3}-\sqrt{2}}-\sqrt{2\sqrt{3}+\sqrt{2}}\right)}{\sqrt{10}}\)
\(=\frac{3}{\sqrt{10}}\)
\(\sqrt{\frac{9}{\sqrt{14+4\sqrt{6}}}}-\sqrt{\frac{9}{\sqrt{14-4\sqrt{6}}}}\)
\(=\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2+2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}-\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2-2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}\)
\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)
\(=\frac{3}{\sqrt{\sqrt{12}+\sqrt{2}}}-\frac{3}{\sqrt{\sqrt{12}-\sqrt{2}}}=\frac{3\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)-3\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)}{\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)}\)
\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\left(\sqrt{\sqrt{12}+\sqrt{2}}\right)\left(\sqrt{\sqrt{12}-\sqrt{2}}\right)}\)
\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{12-2}}\)\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{10}}\)
\(=\frac{3\left(\sqrt{2\sqrt{3}-\sqrt{2}}-\sqrt{2\sqrt{3}+\sqrt{2}}\right)}{\sqrt{10}}\)
bí....!!!
\(\sqrt{3+\sqrt{5}}-\sqrt{4+\sqrt{7}}+\frac{\sqrt{14}}{2}\)
\(\approx1,581\)
A=\(\sqrt{\left(\sqrt{7}-2\right)^2}\)+\(\frac{25\sqrt{7}-63}{3\sqrt{7}-7}\)=\(\frac{12\sqrt{7}-28}{3\sqrt{7}-7}\)=4
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
(đkxđ: \(c\ge0,c\ne4\))
Ta có \(A=\left(\frac{\sqrt{c}}{\sqrt{c}+2}-\frac{\sqrt{c}}{\sqrt{c}-2}+\frac{4\sqrt{c}-1}{c-4}\right).\left(\sqrt{c}+2\right)\)
\(=\frac{\sqrt{c}\left(\sqrt{c}-2\right)-\sqrt{c}\left(\sqrt{c}+2\right)+4\sqrt{c}-1}{\left(\sqrt{c}+2\right)\left(\sqrt{c}-2\right)}\left(\sqrt{c}+2\right)\)
\(=\frac{c-2\sqrt{c}-c-2\sqrt{c}+4\sqrt{c}-1}{\left(\sqrt{c}-2\right)}\)
\(=\frac{1}{2-\sqrt{c}}\)
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)