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Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
Xét hiệu :
\(A-B=2\left(\sqrt{1}-\sqrt{2}\right)+2.\left(\sqrt{3}-\sqrt{4}\right)+...+2\left(\sqrt{19}-\sqrt{20}\right)\)
Mà: \(\sqrt{1}
1. Sửa đề:
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)
2.
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)
