Đáp án:
$a)$
Ta có:
$(3+\sqrt{5})^2=9+6\sqrt{5}+5=14+6\sqrt{5}=14+\sqrt{6^2.5}=14+\sqrt{180}$
$(2\sqrt{2}+\sqrt{6})^2=8+4\sqrt{12}+6=14+\sqrt{4^2.12}=14+\sqrt{192}$
Ta thấy : $14+\sqrt{180}<14+\sqrt{192}⇒(3+\sqrt{5})^2<(2\sqrt{2}+\sqrt{6})^2$
$⇒3+\sqrt{5}<2\sqrt{2}+\sqrt{6}$
$b)$
Ta có:
$(2\sqrt{3}+4)^2=12+16\sqrt{3}+16=28+\sqrt{16^2.3}=28+\sqrt{768}$
$(3\sqrt{2}+\sqrt{10})^2=18+6\sqrt{20}+10=28+\sqrt{6^2.20}=28+\sqrt{720}$
Ta thấy :
$28+\sqrt{768}>28+\sqrt{720}$
$⇒(2\sqrt{3}+4)^2>(3\sqrt{2}+\sqrt{10})^2$
$⇒2\sqrt{3}+4>3\sqrt{2}+\sqrt{10}$
