g) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3^2}}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3^2}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2+\sqrt{3}\right|-\left|2-\sqrt{3}\right|\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
Lời giải:
g.
$\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{(2+\sqrt{3})^2}-\sqrt{(2-\sqrt{3})^2}$
$=|2+\sqrt{3}|-|2-\sqrt{3}|=(2+\sqrt{3})-(2-\sqrt{3})$
$=2\sqrt{3}$
h.
$\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}$
$=\sqrt{(6-\sqrt{5})^2}-\sqrt{(6+\sqrt{5})^2}$
$=|6-\sqrt{5}|-|6+\sqrt{5}|$
$=(6-\sqrt{5})-(6+\sqrt{5})=-2\sqrt{5}$
i.
$=\sqrt{(3+\sqrt{2})^2}+\sqrt{(3-\sqrt{2})^2}$
$=|3+\sqrt{2}|+|3-\sqrt{2}|=6$
j.
$=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|$
$=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
k.
$=\sqrt{20+2\sqrt{20}.\sqrt{4}+4}+\sqrt{4-2\sqrt{4}.\sqrt{5}+5}$
$=\sqrt{(\sqrt{20}+\sqrt{4})^2}+\sqrt{(\sqrt{4}-\sqrt{5})^2}$
$=|\sqrt{20}+\sqrt{4}|+|\sqrt{4}-\sqrt{5}|$
$=\sqrt{20}+\sqrt{4}+\sqrt{5}-\sqrt{4}=\sqrt{20}+\sqrt{5}=3\sqrt{5}$
l.
$=\sqrt{8-2\sqrt{8}.\sqrt{9}+9}+\sqrt{8+2\sqrt{8}.\sqrt{1}+1}$
$=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+1)^2}$
$=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+1|$
$=\sqrt{9}-\sqrt{8}+\sqrt{8}+1=4$
j: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1\)
\(=2\sqrt{3}\)