Bài 1.3:
a) Ta có: \(\left\{{}\begin{matrix}5\sqrt{7}=\sqrt{25.7}=\sqrt{175}\\7\sqrt{5}=\sqrt{49.5}=\sqrt{245}\end{matrix}\right.\)
Do \(245>175\Rightarrow7\sqrt{5}>5\sqrt{7}\Rightarrow\sqrt{7\sqrt{5}}=\sqrt{5\sqrt{7}}\)
b) Ta có: \(\left(\dfrac{\sqrt{5}+1}{\sqrt{2}}\right)^2=\dfrac{6+2\sqrt{5}}{2}=3+\sqrt{5}\)
\(\Rightarrow\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
Bài 1.4:
a) Ta có: \(\sqrt{13}-\sqrt{12}=\dfrac{1}{\sqrt{13}+\sqrt{12}}\); \(\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
Do \(\sqrt{13}+\sqrt{12}>\sqrt{12}+\sqrt{11}\Rightarrow\dfrac{1}{\sqrt{13}+\sqrt{12}}< \dfrac{1}{\sqrt{12}+\sqrt{11}}\Rightarrow\sqrt{13}-\sqrt{12}< \sqrt{12}-\sqrt{11}\)
b) Ta có: \(21+4\sqrt{5}=\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1=\left(2\sqrt{5}+1\right)^2\)
\(\Rightarrow7-\sqrt{21+4\sqrt{5}}=7-\left(2\sqrt{5}+1\right)=6-2\sqrt{5}\)
Lại có: \(\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow7-\sqrt{21+4\sqrt{5}}=\left(\sqrt{5}-1\right)^2\) \(\Rightarrow\sqrt{7-\sqrt{21+4\sqrt{5}}}=\sqrt{5}-1\)
