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21 tháng 5 2022
1: \(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)\)
\(=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)
2: \(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
3: \(=\left(\sqrt{7}-\sqrt{2}+\sqrt{7}+\sqrt{2}\right)^2=\left(2\sqrt{7}\right)^2=28\)
29 tháng 7 2020
b) Ta có: \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(\frac{5+2\sqrt{6}+2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\frac{5+2\sqrt{6}+10-4\sqrt{6}}{25-24}\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=15^2-\left(2\sqrt{6}\right)^2\)
\(=225-24=201\)
12 tháng 12 2018
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
28 tháng 10 2018
\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{\sqrt{9}.\sqrt{2}-\sqrt{4}.\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{(\sqrt{3}-\sqrt{2})\sqrt{6^2}}{\sqrt{3}-\sqrt{2}}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\sqrt{6}-\frac{5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{6+6\sqrt{6}-5}{1+\sqrt{6}}\)
\(\Rightarrow\frac{1+6\sqrt{6}}{1+\sqrt{6}}\)
ON
27 tháng 8 2018
Bạn để ý nhé cách tính là nhân cả tử và mẫu với căn 2
1) \(=\frac{\sqrt{2}.\sqrt{5-\sqrt{21}}}{\sqrt{2}}\)
\(=\frac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\sqrt{21}+3}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
\(=\frac{\sqrt{14}-\sqrt{6}}{2}\)
câu 2 bạn làm tương tự nhé
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
mik sẽ làm chi tiết, không hiểu thì mình thua.
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(1-2\sqrt{3}\right)}}\)
\(=\sqrt{6+2\sqrt{5-1-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(1-\sqrt{3}\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=1+\sqrt{3}\)