\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)
C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$
Ta có: $3>2$
$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$
C2: $3\sqrt{7}=\sqrt{63}$
Ta có: $63>28$
$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$