\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-6\sqrt{20}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

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

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

Khó zậy

\(2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}\)

\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)

\(=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)

\(N=\frac{x^2+2000}{x}=x+\frac{2000}{x}\ge2\sqrt{x.\frac{2000}{x}}=2\sqrt{2000}=40\sqrt{5}\)

Dấu "=" tại \(x=20\sqrt{5}\)

\(Q=\frac{x^2+2x+17}{2\left(x+1\right)}=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{x+1}{2}.\frac{8}{x+1}}=4\)

Dấu "=" tại x = 3 

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

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

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

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

#)Giải :

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

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

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

mn ơi GIÚP E MAI E ĐI HỌC RỒI

\(a,\hept{\begin{cases}x^2+y^2+x+y=4\\x\left(x+y+1\right)+y\left(y+1\right)=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2+y^2+x+y=4\\x^2+xy+x+y^2+y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2+y^2+x+y=4\\x^2+y^2+x+y+xy=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2+y^2+x+y=4\\xy=-2\end{cases}}\)(Trừ 2 pt cho nhau)

\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2+x+y-2xy=4\\xy=-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2+x+y+4=4\\xy=-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)\left(x+y+1\right)=0\\xy=-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=0\\xy=-2\end{cases}\left(h\right)\hept{\begin{cases}x+y+1=0\\xy=-2\end{cases}}}\)