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Giải thích các bước giải:

B1)

a) $\sqrt{4-2\sqrt3}+\sqrt{7+4\sqrt3}$

$=\sqrt{(\sqrt3-1)^2}+\sqrt{\sqrt3+2)^2}$

$=|\sqrt3-1|+|\sqrt3+2|$

$=\sqrt3-1+\sqrt3+2$
$=2\sqrt3+1$

b) $2-\sqrt3-\sqrt{4+2\sqrt3}$
$=2-\sqrt3-\sqrt{(2+\sqrt3)^2}$

$=2-\sqrt3-|2+\sqrt3|$
$=2-\sqrt3-2-\sqrt3$
$=-2\sqrt3$

c) $(\sqrt5+2).\sqrt{9-4\sqrt5}$

$=(\sqrt5+2).\sqrt{(\sqrt5-2)^2}$
$=(\sqrt5+2).(|\sqrt5-2|)$
$=(\sqrt5+2).(\sqrt5-2)$

$=(\sqrt5)^2-2^2$
$=5-4=1$

d) $\frac{2-\sqrt2}{3\sqrt2}$

$=\frac{\sqrt2(\sqrt2-1)}{3\sqrt2}$
$=\frac{\sqrt2-1}3$

B2)
a) $\frac{x-2\sqrt x+1}{2\sqrt x-3}$
$=\frac{(\sqrt x-1)^2}{2\sqrt x-3}$
b) $\frac{x-9}{2\sqrt5+6}$
$=\frac{x-9}{2(\sqrt5+3)}$
c) $\frac{x-1}{\sqrt x-1}:\frac{\sqrt x+1}2$
$=\frac{x-1}{\sqrt x-1}.\frac2{\sqrt x+1}$
$=\frac{2(x-1)}{(\sqrt x-1)(\sqrt x+1)}$
$=\frac{2(x-1)}{x-1}$
$=2$

d) $\frac{x+4\sqrt x+4}{3\sqrt x}:\frac{3\sqrt x+6}{\sqrt x}$
$=\frac{(\sqrt x+2)^2}{3\sqrt x}.\frac{\sqrt x}{3(\sqrt x+2)}$
$=\frac{\sqrt x+2}9$

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