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\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=x\Rightarrow x^2=5x\Rightarrow x=5\)(n số 5)
Vậy \(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=5\) khi \(n\rightarrow\infty\)
\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=x\\ \Leftrightarrow x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}\\ \Leftrightarrow x^2=6+x\Rightarrow x=3\)(n số 6)
\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=3\) khi \(n\rightarrow\infty\)
Vậy S < 8
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
a) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(\sqrt{9\cdot11}-\sqrt{9\cdot2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=3\cdot11-3\sqrt{22}-11+3\sqrt{22}\)
\(=33-11=22\)
b)\(3\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}\)
\(=\frac{9}{\sqrt{8}}-\frac{7}{\sqrt{2}}+\frac{5}{\sqrt{18}}\)
\(=\frac{9}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)
\(=\frac{27-42+10}{6\sqrt{2}}\)
\(=-\frac{5}{6\sqrt{2}}\)
c)\(\left(1+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\frac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(=\left(1-\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}+1\right)\)
\(=\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(=1-5=-4\)
Ta có cái đầu <5
Cái sau <3 nên VT <8
Cảm ơn bạn nhe