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\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{15+2.3.\sqrt{6}}\)\(-\sqrt{10+2.2\sqrt{6}}\)
\(=\sqrt{9+2.3\sqrt{6}+6}\)\(-\sqrt{6+2.\sqrt{6}.2+4}\)
\(=\sqrt{\left(3+\sqrt{6}\right)^2}\)\(-\sqrt{\left(\sqrt{6}+2\right)^2}\)
\(=3+\sqrt{6}\)\(-2\)\(-\sqrt{6}=\left(3-2\right)+\left(\sqrt{6}-\sqrt{6}\right)\)
\(=1+0=1\)
a) \((\sqrt{3}-\sqrt{2}).\sqrt{(\sqrt{3}+\sqrt{2})^2}\)
\(\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)\)
\(\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)\(=3-2=1\)
b) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{(2+2\sqrt{5})^2}+\sqrt{(\sqrt{5}-2)^2}\)
=\(2+2\sqrt{5}+\sqrt{5}-2\)\(=3\sqrt{5}\)
\(13-2\sqrt{42}=7-2\sqrt{42}+6\\ =\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{7}-\sqrt{6}\right)^2\)
\(46+6\sqrt{5}=\left(5+2\cdot\sqrt{5}\cdot3+9\right)+32=\left(\sqrt{5}+3\right)^2+32\)(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 \(x< 0\))
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)