Question

a.\(\sqrt{15+6\sqrt{6}}\)

b.\(\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

c. \(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

Answer 1

Answer 2

\(a,\sqrt{15+6\sqrt{6}}=\sqrt{9+6\sqrt{6}+6}=\left(3+\sqrt{6}\right)^2\)

\(b,Dat:\left\{{}\begin{matrix}\sqrt{17-12\sqrt{2}}=a\\\sqrt{17+12\sqrt{2}}=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab=1\\a^2+b^2=34\end{matrix}\right.\Rightarrow a^2+b^2+2ab=\left(a+b\right)^2=36=\left(\pm6\right)^2.\Rightarrow\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=6\left(\sqrt{17-12\sqrt{2}};\sqrt{17+12\sqrt{2}}\ge0\right)\)

\(c,=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\left(\sqrt{2}>1=\sqrt{1};\sqrt{3}>1=\sqrt{1};2=\sqrt{4}>\sqrt{3}\right)\)