\(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
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6 tháng 8 2021
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
20 tháng 10 2018
a) \(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7\left(\sqrt{3}+\sqrt{5}\right)}}=\) \(\frac{\sqrt{2}}{\sqrt{7}}\)
b ) \(\frac{15\sqrt{2}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=\frac{3\left(5\sqrt{2}+3\sqrt{3}\right)}{3\left(\sqrt{3}+\sqrt{5}\right)}\)\(=\frac{5\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)
c)\(\frac{\sqrt{2}-\sqrt{6}+\sqrt{3}-\sqrt{9}+\sqrt{4}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)+\sqrt{3}\left(1-\sqrt{3}\right)+\sqrt{4}\left(1-\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)\(=\frac{\left(1-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)
d) \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
PN
29 tháng 7 2020
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
7 tháng 9 2020
+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)
\(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)
\(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)
\(=4\sqrt{3}\approx6,9282\)
+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)
\(=\sqrt{x-9+6\sqrt{x-9}+9}\)
\(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)
\(=\left|\sqrt{x-9}-3\right|\)
7 tháng 9 2020
\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)
6 tháng 8 2017
\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
6 tháng 8 2017
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)
6 tháng 8 2020
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@@22@22@22@@222@@2@@2@@@2@2
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)