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\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)
\(\Leftrightarrow10\)
\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-7\)
\(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
\(\dfrac{2-\sqrt{5}}{2+\sqrt{5}}+\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
\(=\dfrac{\left(2-\sqrt{5}\right)^2}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}+\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{4-4\sqrt{5}+5}{4-5}+\dfrac{5+4\sqrt{5}+4}{5-4}\)
\(=-4+4\sqrt{5}-5+5+4\sqrt{5}+4\)
\(=8\sqrt{5}\)
\(=\dfrac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}+\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{9-4\sqrt{5}}{-1}+9+4\sqrt{5}\)
=9+4căn 5-9+4căn 5
=8*căn 5
Ta có: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\sqrt{5}+2-\sqrt{5}-2=0\)
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
A = 2 ( 3 + 5 ) 2 2 + 3 + 5 + 2 ( 3 − 5 ) 2 2 − 3 − 5 2 3 + 5 4 + ( 5 + 1 ) 2 + 3 − 5 4 − ( 5 − 1 ) 2 = 2 3 + 5 5 + 5 + 3 − 5 5 − 5 2 ( 3 + 5 ) ( 5 − 5 ) + ( 3 − 5 ) ( 5 + 5 ) ( 5 + 5 ) ( 5 − 5 ) = 2 15 − 3 5 + 5 5 − 5 + 15 + 3 5 − 5 5 − 5 25 − 5 = 2. 20 20 = 2 V ậ y A = 2
