Lời giải:
a.
$4\sqrt{3}=\sqrt{48}<\sqrt{49}$ hay $4\sqrt{3}< 7$
b.
$-4\sqrt{17}=-\sqrt{16}.\sqrt{17}>-\sqrt{17}.\sqrt{17}$ hay $-4\sqrt{17}>-17$
c.
$7+\sqrt{14}<7+\sqrt{16}$ hay $7+\sqrt{14}< 11$
d.
$9-\sqrt{5}-7=2-\sqrt{5}<2-\sqrt{4}=0$
$\Rightarrow 9-\sqrt{5}< 7$
e.
$(\sqrt{5}+2)^2=9+4\sqrt{5}=9+2\sqrt{20}$
$(\sqrt{2}+\sqrt{6})^2=8+2\sqrt{12}$
Dễ thấy $9+2\sqrt{20}>8+2\sqrt{12}$ nên $(\sqrt{5}+2)^2>(\sqrt{2}+\sqrt{6})^2$
$\Rightarrow \sqrt{5}+2>\sqrt{2}+\sqrt{6}$
f.
$(\sqrt{3}+2)^2=7+4\sqrt{3}$
$(\sqrt{2}+\sqrt{6})^2=8+2\sqrt{12}=8+4\sqrt{3}$
$\Rightarrow (\sqrt{3}+2)^2< (\sqrt{2}+\sqrt{6})^2$
$\Rightarrow \sqrt{3}+2<\sqrt{2}+\sqrt{6}$
$\Rightarrow -\sqrt{3}-2>-\sqrt{2}-\sqrt{6}$
g.
$3-2\sqrt{2}-2=1-2\sqrt{2}<0$
$\Rightarrow 3-2\sqrt{2}<2$
h.
$\frac{7}{3}\sqrt{\frac{1}{12}}=\frac{7}{3}.\sqrt{\frac{1}{2^2.3}}$
$=\frac{7}{3}.\frac{1}{2\sqrt{3}}=\frac{7}{6\sqrt{3}}< \frac{7}{4\sqrt{3}}$
