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\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}\right)^2+\frac{1}{2}\left(\sqrt{5}\right)^2=\frac{15}{2}+\frac{5}{2}=\frac{20}{2}=10\)
\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3.\sqrt{5}}-\sqrt{2}\)
\(\sqrt{2}.A=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{9-2.3.\sqrt{5}+5}-2\)
\(\sqrt{2}.A=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow A=\sqrt{2}\)
ĐKXĐ: \(\hept{\begin{cases}2x-4\ge0\\x+2.\sqrt{2x-4}\ge0\\x-2\sqrt{2x-4}\end{cases}}\Leftrightarrow x\ge2\)
\(\sqrt{x+2.\sqrt{2x-4}}+\sqrt{x-2.\sqrt{2x-4}}\)
\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)
\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Tự phá trị tuyệt đối
\(A=\sqrt{\left(3+\sqrt{3}\right)+2\sqrt{\left(3+\sqrt{3}\right)\left(\sqrt{5}-2\right)+\left(\sqrt{5}-2\right)}-\sqrt{3+\sqrt{3}}}\)
\(=\sqrt{3+\sqrt{3}}+\sqrt{\sqrt{5}-2}-\sqrt{3+\sqrt{3}}=\sqrt{\sqrt{5}-2}\)
ok???
A = \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)
<=> A2 = ( \(\sqrt{7+3\sqrt{5}}+\sqrt{7-3\sqrt{5}}\) )2
= 7 + \(3\sqrt{5}\) + \(2\sqrt{\left(7+3\sqrt{5}\right).\left(7-3\sqrt{5}\right)}\) + 7 - \(3\sqrt{5}\)
= 14 + 2\(\sqrt{7^2-\left(3\sqrt{5}\right)^2}\)
= 14 + 2\(\sqrt{4}\)
= 18
=> A = \(\sqrt{18}\)
B = \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\)
<=> B2 = ( \(\sqrt{2-\sqrt{2\sqrt{5}-2}}\) - \(\sqrt{2+\sqrt{2\sqrt{5}-2}}\) )2
= 2 - \(\sqrt{2-2\sqrt{5}}\) - 2\(\sqrt{\left(2-\sqrt{2\sqrt{5}-2}\right)\left(2+\sqrt{2\sqrt{5}-2}\right)}\) + 2 + \(\sqrt{2-2\sqrt{5}}\)
= 4 - 2 \(\sqrt{2^2-\left(\sqrt{2\sqrt{5}-2}\right)^2}\)
= 4 - 2 \(\sqrt{4-\left(2\sqrt{5}-2\right)}\)
= 4 - 2 \(\sqrt{4-2\sqrt{5}+2}\)
= 4 - 2 \(\sqrt{\left(\sqrt{5}-1\right)^2}\)
= 4 - 2( \(\sqrt{5}-1\) )
= 6 - 2\(\sqrt{5}\)
=> B = \(\sqrt{6-2\sqrt{5}}\) = \(\sqrt{\left(\sqrt{5}-1\right)^2}\) = \(\sqrt{5}-1\)
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
\(A=\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}+\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{8-2\sqrt{7}}\)
\(A=\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}-2\sqrt{7}+2\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(B=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{2^2}\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\sqrt[3]{2}\)