tính :
A=\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
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15 tháng 7 2016
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
15 tháng 7 2016
\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
LT
1
27 tháng 9 2020
Ta có: \(\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2-2\sqrt{2}\cdot1+1}}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{\left(\sqrt{2}-1\right)^2}}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\left|\sqrt{2}-1\right|}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\left(\sqrt{2}-1\right)}}\)(Vì \(\sqrt{2}>1\))
\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2}+8}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{18-8\sqrt{2}}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}\)
\(=\sqrt{4\sqrt{2}+4\sqrt{\left(4-\sqrt{2}\right)^2}}\)
\(=\sqrt{4\sqrt{2}+4\left|4-\sqrt{2}\right|}\)
\(=\sqrt{4\sqrt{2}+4\left(4-\sqrt{2}\right)}\)(Vì \(4>\sqrt{2}\))
\(=\sqrt{4\sqrt{2}+16-4\sqrt{2}}\)
\(=\sqrt{16}=4\)
TN
2
AT
4 tháng 7 2018
\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{8}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}=\sqrt{4+\sqrt{8}}\cdot\sqrt{4-2-\sqrt{2}}=\sqrt{4+\sqrt{8}}\cdot\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}=\sqrt{8+\sqrt{32}-\sqrt{32}-4}=\sqrt{4}=2\)
4 tháng 7 2018
\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2\left(4-2\right)}=\sqrt{2.2}=2\)
23 tháng 6 2015
\(y=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(y=1-\frac{1}{10}=\frac{9}{10}\)