Bài 1:
\(\dfrac{\sqrt{6}+\sqrt{8}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{6}+2\sqrt{2}}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\left(\sqrt{6}+2\sqrt{2}\right)\left(2\sqrt{3}-2\sqrt{7}\right)}{-16}\)
\(=\dfrac{\left(\sqrt{6}+2\sqrt{2}\right)\cdot2\left(\sqrt{3}-\sqrt{7}\right)}{-16}\)
\(=\dfrac{\left(\sqrt{6}+2\sqrt{2}\right)\left(\sqrt{3}-\sqrt{7}\right)}{-8}\)
\(=\dfrac{\sqrt{18}-\sqrt{42}+2\sqrt{6}-2\sqrt{14}}{-8}\)
\(=\dfrac{3\sqrt{2}-\sqrt{42}+2\sqrt{6}-2\sqrt{14}}{-8}\)
\(=-\dfrac{3\sqrt{2}-\sqrt{42}+2\sqrt{6}-2\sqrt{14}}{8}\)
Bài 2 :
\(\left(\sqrt{2+\sqrt{3}}\right)^2=2+\sqrt{3}\)
\((\sqrt{10})^2=10\)\(=2+8\)
\(\)\(3>\sqrt{3};3< 8=>\sqrt{3}< 8\)
\(=>2+\sqrt{3}< 10\)
\(=>\sqrt{2+\sqrt{3}}< \sqrt{10}\)
\(=>Đpcm\)
