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\(13-2\sqrt{42}=7-2\sqrt{42}+6\\ =\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{7}-\sqrt{6}\right)^2\)
\(46+6\sqrt{5}=\left(5+2\cdot\sqrt{5}\cdot3+9\right)+32=\left(\sqrt{5}+3\right)^2+32\)(ko rút đc)
\(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\\ =\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{5-2\sqrt{5}+1}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\\ =4\left(3+\sqrt{5}\right)\)
\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Dễ dàng nhận ra
\(\sqrt{\sqrt{7}-\sqrt{3}}< \sqrt{\sqrt{7}+\sqrt{3}}\Rightarrow\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}< 0\)
Đặt \(x=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}< 0\)
\(\Rightarrow x^2=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}}{\sqrt{7}-2}\)
\(\Rightarrow x^2=\frac{2\sqrt{7}-2\sqrt{4}}{\sqrt{7}-2}=\frac{2\sqrt{7}-4}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
\(\Rightarrow x=-\sqrt{2}\) (do \(x< 0\))
a) \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2\sqrt{3}+2-\sqrt{3}\)
\(=\left(2\sqrt{3}-\sqrt{3}\right)+2\)
\(=\sqrt{3}+2\)
b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{1+\sqrt{5}}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{1+\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{12}{4}=3\)
c) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{1}=14\)
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{1}=1}\)
b) \(B=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{21-6\sqrt{12}}}=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{\left(3-2\sqrt{3}\right)^2}}}}=\sqrt{\sqrt{3}-\sqrt{2\sqrt{3}-2}}\)c)
\(C=\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{2+2\sqrt{5}}{\sqrt{2}}=\sqrt{2}+\sqrt{10}=\sqrt{2}\left(\sqrt{5}+1\right)\)
\(a)\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}-1=1\)
\(b)\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{33-2.3.\sqrt{4}.\sqrt{6}}\)
\(=3-\sqrt{6}+\sqrt{33-2.3.\sqrt{24}}\)
\(=3-\sqrt{6}+\sqrt{\left(\sqrt{24}-3\right)^2}\)
\(=3-\sqrt{6}+\sqrt{24}-3\)
\(=\sqrt{24}-\sqrt{6}\)
\(=\sqrt{6}\left(2-1\right)=\sqrt{6}\)
\(c)\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}+\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(=\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}+\sqrt{\frac{\left(3+\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}\)
\(=\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{4}}+\sqrt{\frac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\frac{3-\sqrt{5}}{2}+\frac{3+\sqrt{5}}{2}\)
\(=\frac{6}{2}=3\)
\(d)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\frac{24}{2}=12\)
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. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)