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13 tháng 7 2018
a,\(\sqrt{2\left(11+6\sqrt{2}\right)}\)=\(\sqrt{2\left(9+2.3.\sqrt{2}+2\right)}\)=\(\sqrt{2\left(3+\sqrt{2}\right)^2}\)=\(\sqrt{2}\)(3+\(\sqrt{2}\))
13 tháng 7 2018
\(a.\sqrt{22+12\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.2+4}=3\sqrt{2}+2\)
\(b.\sqrt{\dfrac{5+2\sqrt{6}}{2}}=\sqrt{\dfrac{3+2\sqrt{3}.\sqrt{2}+2}{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{2}\)
\(c.\sqrt{30+4\sqrt{2}.\sqrt{7}}=\sqrt{28+2.\sqrt{2}.2\sqrt{7}+2}=2\sqrt{7}+\sqrt{2}\)
\(d.\sqrt{5+2\sqrt{2-\sqrt{9-4\sqrt{2}}}}=\sqrt{5+2\sqrt{2-\sqrt{8-2.2\sqrt{2}+1}}}=\sqrt{5+2\sqrt{2-2\sqrt{2}+1}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1\) \(e.\sqrt{1+2\sqrt{\sqrt{2+\sqrt{11+6\sqrt{2}}}}}=\sqrt{1+2\sqrt{\sqrt{2+\sqrt{9+2.3\sqrt{2}+2}}}}=\sqrt{1+2\sqrt{\sqrt{5+\sqrt{2}}}}\)
\(f.\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{1-\dfrac{\sqrt{3}}{2}}}=\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{\dfrac{3}{4}-2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}}=\sqrt{\sqrt{3}+\dfrac{1}{2}}=\)
\(g.\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}+\sqrt{3+2\sqrt{3}+1}=\sqrt{7}-\sqrt{3}+\sqrt{3}+1=\sqrt{7}+1\)
12 tháng 8 2020
1) Cách 1 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(M=\sqrt{9-6\sqrt{2}+2}+\sqrt{9+6\sqrt{2}+2}\)
\(M=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(M=\left|3-\sqrt{2}\right|+\left|3+\sqrt{2}\right|\)
\(M=3-\sqrt{2}+3+\sqrt{2}=6\)
Cách 2 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(\Rightarrow M^2=11-6\sqrt{2}+2\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}+11+6\sqrt{2}\)
\(\Leftrightarrow M^2=22+2.7=36\)
\(\Leftrightarrow M=6\left(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}>0\right)\)
2)
\(A=53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{8-4\sqrt{2}+1}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\left|2\sqrt{2}-1\right|}\)
\(\Leftrightarrow A=53-20\sqrt{4+2\sqrt{2}-1}\)
\(\Leftrightarrow A=53-20\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow A=53-20\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow A=53-20\left(\sqrt{2}+1\right)\)
\(\Leftrightarrow A=53-20\sqrt{2}-20=33-20\sqrt{2}\)
12 tháng 8 2020
3)
\(M=\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(M=\sqrt{3-\sqrt{5}}.\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\)
\(M=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(M=2\sqrt{2}.\sqrt{3-\sqrt{5}}\left(\sqrt{5}+1\right)\)
\(\Rightarrow M^2=8.\left(3-\sqrt{5}\right).\left(5+2\sqrt{5}+1\right)\)
\(\Leftrightarrow M^2=\left(24-8\sqrt{5}\right)\left(6+2\sqrt{5}\right)\)
\(\Leftrightarrow M^2=144+48\sqrt{5}-48\sqrt{5}-80\)
\(\Leftrightarrow M^2=64\Leftrightarrow M=8\left(\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)>0\right)\)
\(\sqrt{11+2\sqrt{30}}+\sqrt{9+2\sqrt{20}}=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{6}+\sqrt{5}+\sqrt{5}+2=\sqrt{6}+2\sqrt{5}+2\)