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\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)=\(\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)
=\(\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)=\(\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\sqrt{13+30\sqrt{2}+30}\)
=\(\sqrt{43+30\sqrt{2}}\)=\(\sqrt{\left(5+3\sqrt{2}\right)^2}\)=\(5+3\sqrt{2}\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}.\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{13+30\sqrt{2}+30}=\sqrt{43+30\sqrt{2}}\)
Giải từ từ lần lượt các biểu thức trong dấu căn nhé:
\(\sqrt{13+\sqrt{48}}=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}=\sqrt{\left(2\sqrt{3}+1\right)^2}=2\sqrt{3}+1\)
\(\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(B=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{3+2\sqrt{3}+1}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+12}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-1-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{1-2\sqrt{3}+\sqrt{3}^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(1-\sqrt{3}\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{6-2}\)
\(\frac{\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(\sqrt{13+\sqrt{30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}}\) \(=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}}=\sqrt{13+\sqrt{30\sqrt{2+2\sqrt{2}+1}}}\)\(=\sqrt{13+\sqrt{30\sqrt{\left(\sqrt{2}+1\right)^2}}}\)\(=\sqrt{13+\sqrt{30\left(\sqrt{2}+1\right)}}=\sqrt{13+\sqrt{30\sqrt{2}+30}}\)