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\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\sqrt{9-5}\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)
\(=2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=2\left(5-1\right)\)
\(=8\)
Ta có : \(94-42\sqrt{5}=45-2.7.3\sqrt{5}+49=\left(3\sqrt{5}\right)^2-2.7.3\sqrt{5}+7^2=\left(7-3\sqrt{5}\right)^2\)
\(94+42\sqrt{5}=\left(7+3\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
d) \(\frac{1}{\sqrt{3}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}-\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{2\sqrt{5}}{3-5}=\frac{2\sqrt{5}}{-2}=-\sqrt{5}\)c) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}+\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
b) \(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}+\sqrt{5-2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)a) \(\sqrt{27}+\sqrt{243}-6\sqrt{12}=\sqrt{9.3}+\sqrt{81.3}-6\sqrt{4.3}=3\sqrt{3}+9\sqrt{3}-12\sqrt{3}=0\)
\(a.\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.2\sqrt{2}.3\sqrt{5}+8}=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}=-4\sqrt{5}\) \(b.\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}.\sqrt{3}+3}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)
\(\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{6-2\sqrt{5}}\)
\(=3-\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=3-\sqrt{5}+\sqrt{5}-1=2\)
\(\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}+2-\sqrt{5}=2\)
Chúc học tốt!!!!!!!!!!!!!
\(\left(4+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)