Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
1, \(\sqrt{19+8\sqrt{3}}-\sqrt{28-6\sqrt{3}}+\sqrt{12}\)
\(=\sqrt{4^2+2.4\sqrt{3}+3}-\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}+1}+2\sqrt{3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{3}-1\right)^2}+2\sqrt{3}\)
\(=4+\sqrt{3}-3\sqrt{3}+1+2\sqrt{3}=5\)
2, \(\left(2+\frac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\frac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(=\left(2+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2-\sqrt{5}}\right)\left(2+\frac{\sqrt{5}\left(\sqrt{5}+3\right)}{3+\sqrt{5}}\right)\)
\(=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1\)
3, \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}-\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2}}+\sqrt{3^2-2.3\sqrt{6}+6}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}-\left(5-2\sqrt{6}\right)+3-\sqrt{6}\)
\(=4\sqrt{6}+3-\sqrt{6}=3+3\sqrt{6}\)