Tính:
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
K
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Những câu hỏi liên quan
21 tháng 4 2017
Ta có:
\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)
\(\Leftrightarrow A^2=10-2\sqrt{25-17}=10-4\sqrt{2}\)
\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)
Ta lại có:
\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)
\(\Leftrightarrow B=\sqrt{2}\)
Thế vô biểu thức ban đầu ta được
\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)
KT
6 tháng 7 2018
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
KT
6 tháng 7 2018
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
12 tháng 7 2018
\(1.\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=\sqrt{17-4\left(\sqrt{5}+2\right)}=\sqrt{5-2.2\sqrt{5}+4}=\sqrt{5}-2\)
\(2.\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}=\sqrt{17-6\left(\sqrt{2}+1\right)}=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)\(3.\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{3+\sqrt{3-2\sqrt{3}+1}}=\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(4.\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}=\sqrt{25+2.5\sqrt{2}+2}.\left(5-\sqrt{2}\right)=\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)=5-2=3\)
27 tháng 7 2021
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
18 tháng 6 2021
`2+\sqrt{17-4sqrt{9+4sqrt5}}`
`=2+sqrt{17-4sqrt{4+2.2sqrt5+5}}`
`=2+sqrt{17-4sqrt{(sqrt5+2)^2}}`
`=2+sqrt{17-4(sqrt5+2)}`
`=2+sqrt{9-4sqrt5}`
`=2+sqrt{5-2.2sqrt5+4}`
`=2+sqrt{(sqrt5-2)^2}`
`=2+sqrt5-2=sqrt5`
NM
1
27 tháng 6 2017
\(\sqrt{10+2\sqrt{17-4\left(2+\sqrt{5}\right)}}\)=\(\sqrt{10+2\sqrt{17-8-4\sqrt{5}}}=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
=\(\sqrt{10+2\left(\sqrt{5}-2\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
TD
17 tháng 12 2016
a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)
b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)
d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)
C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\cdot\left|2+\sqrt{5}\right|}}\)
\(=\sqrt{10+2\sqrt{17-4\cdot\left(2+\sqrt{5}\right)}}\)
\(=\sqrt{10+2\sqrt{17-8-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)
\(=\sqrt{10+2\sqrt{\left(2-\sqrt{5}\right)^2}}\)
\(=\sqrt{10+2\cdot\left|2-\sqrt{5}\right|}\)
\(=\sqrt{10+2\cdot\left(-2+\sqrt{5}\right)}\)
\(=\sqrt{10+-4+2\sqrt{5}}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|\sqrt{5}+1\right|\)
\(=\sqrt{5}+1\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{5+4\sqrt{5}+4}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\cdot\left|\sqrt{5}+2\right|}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\) (vì \(\sqrt{5}+2>0\))
\(=\sqrt{10+2\sqrt{17-4\sqrt{5}-8}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\\ =\sqrt{10+2\sqrt{5-4\sqrt{5}+4}}\\ =\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\\ =\sqrt{10+2\cdot\left|\sqrt{5}-2\right|}\)
\(=\sqrt{10+2\cdot\left(\sqrt{5}-2\right)}\) (vì \(\sqrt{5}-2>0\))
\(=\sqrt{10+2\sqrt{5}-4}\\ =\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}\\ =\left|\sqrt{5}+1\right|\)
\(=\sqrt{5}+1\) (vì \(\sqrt{5}+1>0\))