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26, đặt bthuc là A suy ra A2=4+4+2\(\sqrt{16-\left(10+2\sqrt{5}\right)}\) suy ra A2=8+2(\(\sqrt{5}\) -1) suy ra A=\(\sqrt{6+2\sqrt{5}}\)=\(\sqrt{5}\)+1
40, tương tự
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a) \(E=\sqrt{\left|12\sqrt{5}-29\right|}-\sqrt{12\sqrt{5}+29}\)
\(\Leftrightarrow E^2=\left|12\sqrt{5}-29\right|-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=29-12\sqrt{5}-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=-24\sqrt{5}\)
\(\Leftrightarrow E=-2\sqrt{6\sqrt{5}}\)
b) Đặt \(F=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
\(\Leftrightarrow F^2=\left|40\sqrt{2}-57\right|-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=57-40\sqrt{2}-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=-80\sqrt{2}\)
\(\Leftrightarrow F=-4\sqrt{5\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(L=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{\left|40\sqrt{2}-57\right|}\)
\(=\sqrt{40\sqrt{2}-57}-\sqrt{40\sqrt{2}-57}\)
\(=0\)
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\(\left(2\sqrt{5}.\sqrt{2}-3\sqrt{40}+\sqrt{90}:3\right):\sqrt{640}=\left(2\sqrt{10}-6\sqrt{10}+\sqrt{10}\right):\left(8\sqrt{10}\right)\)
\(=-\frac{3\sqrt{10}}{8\sqrt{10}}=-\frac{3}{8}\)
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1) \(\sqrt{21+12\sqrt{3}}=\sqrt{3^2+2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}=\sqrt{\left(3+2\sqrt{3}\right)^2}\)
\(=\left|3+2\sqrt{3}\right|=3+2\sqrt{3}\)
2) \(\sqrt{57-40\sqrt{2}}=\sqrt{5^2-2.5.4\sqrt{2}+\left(4\sqrt{2}\right)^2}=\sqrt{\left(5-4\sqrt{2}\right)^2}\)
\(=\left|5-4\sqrt{2}\right|=4\sqrt{2}-5\)
3) \(\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left|\sqrt{5}+1\right|+\left|\sqrt{5}-1\right|\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
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Giải:
1) \(\sqrt{21+12\sqrt{3}}\)
\(=\sqrt{12+9+12\sqrt{3}}\)
\(=\sqrt{12+12\sqrt{3}+9}\)
\(=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}.3+3^2}\)
\(=\sqrt{\left(2\sqrt{3}+3\right)^2}\)
\(=2\sqrt{3}+3\)
Vậy ...
2) \(\sqrt{57-40\sqrt{2}}\)
\(=\sqrt{32+25-40\sqrt{2}}\)
\(=\sqrt{32-40\sqrt{2}+25}\)
\(=\sqrt{\left(4\sqrt{2}\right)^2-2.4\sqrt{2}.5+5^2}\)
\(=\sqrt{\left(4\sqrt{2}-5\right)^2}\)
\(=4\sqrt{2}-5\)
Vậy ...
3) \(\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
Vậy ...