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14 tháng 7 2019
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
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8 tháng 8 2022
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
28 tháng 7 2016
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}-1}\)
\(=\sqrt{4-\sqrt{5}}\)
c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3-2=1\)
d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3+1}\)
\(=2\sqrt{3}\)
S
31 tháng 7 2019
\(\sqrt{13-2\sqrt{42}}=\sqrt{6-2\sqrt{6}.\sqrt{7}+7}=\sqrt{\left(\sqrt{6}-\sqrt{7}\right)^2}=\left|\sqrt{6}-\sqrt{7}\right|=\sqrt{7}-\sqrt{6}\)
\(\sqrt{46+6\sqrt{5}}=\sqrt{45+6\sqrt{5}+1}=\sqrt{3^2.5+6\sqrt{5}+1}=\sqrt{3^2.5+2.3.\sqrt{5}+1^2}=\sqrt{\left(3.\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)
\(\sqrt{12-3\sqrt{15}}=\sqrt{3}\sqrt{4-\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{8-2\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{3-2\sqrt{15}+5}=\sqrt{\frac{3}{2}}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\frac{3}{2}}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{15}+5}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{3}\sqrt{5}+5}-\sqrt{3+2\sqrt{3}\sqrt{5}+5}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}=-2\sqrt{3}\)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{5}-\sqrt{5}+1\right)=\sqrt{\frac{1}{2}}.2=\sqrt{\frac{4}{2}}=\sqrt{2}\)
5 tháng 9 2020
a) Ta có: \(A=2\sqrt{2+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12}+1}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3+1-2\sqrt{3}}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\Leftrightarrow A=2\sqrt{2+\sqrt{3}-1}\)
\(\Leftrightarrow A=2\sqrt{\sqrt{3}+1}\)
\(\Leftrightarrow A\approx3,30578\)
b) Ta có: \(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(\Leftrightarrow B=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(\Leftrightarrow B=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(\Leftrightarrow B=\sqrt{2}.\left(4-2\right)\)
\(\Leftrightarrow B=2\sqrt{2}\)
\(\Leftrightarrow B\approx2,82843\)
AH
Akai Haruma
Giáo viên
4 tháng 10 2018
Lời giải:
\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{\frac{6+2\sqrt{5}}{2}}+\sqrt{\frac{6-2\sqrt{5}}{2}}\)
\(\sqrt{\frac{5+1+2\sqrt{5}}{2}}+\sqrt{\frac{5+1-2\sqrt{5}}{2}}=\sqrt{\frac{(\sqrt{5}+1)^2}{2}}+\sqrt{\frac{(\sqrt{5}-1)^2}{2}}\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{10}\)
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\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{20+3-2\sqrt{20.3}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{20}-\sqrt{3})^2}\)
\(=\sqrt{5}-\sqrt{3}-(\sqrt{20}-\sqrt{3})=\sqrt{5}-\sqrt{20}=\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)
AH
Akai Haruma
Giáo viên
4 tháng 10 2018
\(9+4\sqrt{2}=9+2\sqrt{8}=8+1+2\sqrt{8}=(\sqrt{8}+1)^2\)
\(\Rightarrow \sqrt{9+4\sqrt{2}}=\sqrt{8}+1=2\sqrt{2}+1\)
\(\Rightarrow 2+\sqrt{9+4\sqrt{2}}=3+2\sqrt{2}=2+1+2\sqrt{2}=(\sqrt{2}+1)^2\)
\(\Rightarrow \sqrt{2+\sqrt{9+4\sqrt{2}}}=\sqrt{2}+1\)
\(\Rightarrow 13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}=13+30(\sqrt{2}+1)=43+30\sqrt{2}\)
\(\Rightarrow \sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{43+30\sqrt{2}}\)
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\((\sqrt{3}-\sqrt{2})\sqrt{5+2\sqrt{6}}=(\sqrt{3}-\sqrt{2})\sqrt{2+3+2\sqrt{2.3}}\)
\(=(\sqrt{3}-\sqrt{2})\sqrt{(\sqrt{2}+\sqrt{3})^2}\)
\(=(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1\)
\(\sqrt{3}+2\sqrt{13}+\sqrt{8}+\sqrt{5}=14,00764846\)