Đáp án:
$a)2\sqrt{2}+\sqrt{6}>3+\sqrt{5}\\ b) 2\sqrt{3}+4>3\sqrt{2}$
Giải thích các bước giải:
$a)3+\sqrt{5} \text{ và } 2\sqrt{2}+\sqrt{6}\\ (3+\sqrt{5})^2=3^2+2.3\sqrt{5}+5=14+2.3\sqrt{5}=14+2\sqrt{45}\\ (2\sqrt{2}+\sqrt{6})^2=(2\sqrt{2})^2+2.2\sqrt{2}.\sqrt{6}+6=14+2.2\sqrt{2}.\sqrt{6}=14+2\sqrt{48}\\ (2\sqrt{2}+\sqrt{6})^2>(3+\sqrt{5})^2\\ \Rightarrow 2\sqrt{2}+\sqrt{6}>3+\sqrt{5}\\ b)2\sqrt{3}+4 \text{ và } 3\sqrt{2}+\sqrt{10}\\ (2\sqrt{3}+4)^2=(2\sqrt{3})^2+2.2\sqrt{3}.4+4^2=28+2.2\sqrt{3}.4=28+2\sqrt{192}\\ (3\sqrt{2}+\sqrt{10})^2=(3\sqrt{2})^2+2.3\sqrt{2}.\sqrt{10}+10=28+2.3\sqrt{2}.\sqrt{10}=28+2\sqrt{180}\\ (2\sqrt{3}+4)^2>(3\sqrt{2})^2\\ \Rightarrow 2\sqrt{3}+4>3\sqrt{2}$
