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MN
1
9 tháng 6 2017
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-\sqrt{3}^2}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
10 tháng 7 2017
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
TN
1
12 tháng 4 2019
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{20+4\sqrt{3}-10\sqrt{3}-6}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{4\left(5+\sqrt{3}\right)-2\sqrt{3}\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(4-2\sqrt{3}\right)\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\Rightarrow A=2\)
VP
29 tháng 5 2018
mình không viết lại đề nha
\(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2.\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(3+2\sqrt{3}+1\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(4+2\sqrt{3}\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{56-24\sqrt{3}+28\sqrt{3}-36}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{20+4\sqrt{3}}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(20+4\sqrt{3}\right).\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right).\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{100-20\sqrt{3}+20\sqrt{3}-12}{5^2-\sqrt{3}^2}}\)
\(=\sqrt{\frac{88}{25-3}}\)
\(=\sqrt{\frac{88}{22}}\)
\(=\sqrt{4}\)
\(=2\)
HỌC TỐT !!!
EM
3
12 tháng 6 2018
\(A=\frac{\left(\sqrt{3}+1\right)\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}=\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}=\frac{44\left(\sqrt{3}-1\right)}{22}=2\sqrt{3}-2\)
DS
1
28 tháng 7 2020
Sửa đề: \(D=\left(\sqrt{14}-\sqrt{6}\right)\cdot\left(5+\sqrt{21}\right)\cdot\sqrt{4-\sqrt{7}}\)
Ta có: \(D=\left(\sqrt{14}-\sqrt{6}\right)\cdot\left(5+\sqrt{21}\right)\cdot\sqrt{4-\sqrt{7}}\)
\(=\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(5+\sqrt{21}\right)\cdot\sqrt{4-\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}\cdot\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(5+\sqrt{21}\right)\)
\(=\left(\sqrt{7}-1\right)\cdot\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(5+\sqrt{21}\right)\)
\(=\left(7-\sqrt{21}-\sqrt{7}+\sqrt{3}\right)\cdot\left(5+\sqrt{21}\right)\)
\(=35+7\sqrt{21}-5\sqrt{21}-21-5\sqrt{7}-7\sqrt{3}+5\sqrt{3}+3\sqrt{7}\)
\(=14+2\sqrt{21}-2\sqrt{7}-2\sqrt{3}\)
\(\sqrt{\left(\sqrt{5}+3\right)^2}+\sqrt{14-6\sqrt{5}}\)\(=\left|\sqrt{5}+3\right|+\sqrt{9-2.3\sqrt{5}+5}\)
\(=\sqrt{5}+3+\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}+3+\left|3-\sqrt{5}\right|\)
\(=\sqrt{5}+3+3-\sqrt{5}=6\) ( do \(3-\sqrt{5}>0\))