Cách 1: Theo casio ta có:
+ \(\sqrt{3}+\sqrt{7}\approx4,378\)
+ \(\sqrt{19}\approx4,36\)
=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)
Cách 2: Ta có: \(\left(\sqrt{3}+\sqrt{7}\right)^2=3+7+2.\sqrt{21}=10+\sqrt{84}\)
\(\left(\sqrt{19}\right)^2=19=10+\sqrt{81}\)
Vì \(10+\sqrt{84}>10+\sqrt{81}\)
=> \(\left(\sqrt{3}+\sqrt{7}\right)^2>\left(\sqrt{19}\right)^2\)
=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)
Ta có: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}>10+2\sqrt{20,25}=10+2\sqrt{\left(4,5\right)^2}=10+2.4,5=10+9=19=\left(\sqrt{19}\right)^2\)
(Vì 21 > 20,25 > 0 => \(\sqrt{21}>\sqrt{20,25}\))
Mà 2 biểu thức so sánh đều dương
=>\(\sqrt{3}+\sqrt{7}>\sqrt{19}\).
