Đặt \(A=\sqrt{9+\sqrt{13}}+\sqrt{9-\sqrt{13}}\)

\(\Rightarrow A^2=9+\sqrt{13}+9-\sqrt{13}+2\sqrt{\left(9+\sqrt{13}\right)\left(9-\sqrt{13}\right)}\)

\(\Rightarrow A^2=18-2\sqrt{9^2-13}\)

\(\Rightarrow A^2=18-4\sqrt{17}\)

\(\Rightarrow A=\sqrt{18-4\sqrt{17}}\)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{13+30\sqrt{2}+30}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)

#)Giải :

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{8+2.2\sqrt{2+1}}}}}\)

\(=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}}=\sqrt{13+\sqrt{30\sqrt{2+2\sqrt{2+1}}}}\)

\(=\sqrt{13+\sqrt{30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+\sqrt{30\left(\sqrt{2}+1\right)}}=\sqrt{13+\sqrt{30\sqrt{2}+30}}\)

Ta có:

\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}\)

\(=\frac{\sqrt{n+4}-\sqrt{n}}{4\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)

Áp dụng vào bài toán ta được

\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)

\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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\(\sqrt{13+\sqrt{30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}}\) \(=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}}=\sqrt{13+\sqrt{30\sqrt{2+2\sqrt{2}+1}}}\)\(=\sqrt{13+\sqrt{30\sqrt{\left(\sqrt{2}+1\right)^2}}}\)\(=\sqrt{13+\sqrt{30\left(\sqrt{2}+1\right)}}=\sqrt{13+\sqrt{30\sqrt{2}+30}}\)

\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)

\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)

\(=\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}+1\)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

\(M=\frac{3^6.3^8.5^4-3^{15}.5^{13-9}}{3^{12}.5^6+3^{18}.5^9}.2.\frac{25}{9}=\frac{3^{14}.5^4\left(1-3\right).2.5^2}{3^{12}.5^6\left(1+3^6.5^3\right).3^2}=\frac{-4}{1+3^6.5^3}\)

\(\frac{9+4\sqrt{2}}{21}\)

cho  P = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)  , Tìm GTLN của P  

ĐKXĐ: \(x\ge0\)

Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)

Để P lớn nhất thì: \(\frac{1}{\sqrt{x}+1}\)phải lớn nhất.Hay: \(\sqrt{x}+1\)nhỏ nhất

Theo ĐKXĐ,lại có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\)

=>Min \(\sqrt{x}+1\)là 1 tại \(\sqrt{x}=0\Rightarrow x=0\)

=>Max P = \(1+\frac{1}{0+1}=2\)tại x=0

=.= hk tốt!!