Đáp án:
b) $B=6+\sqrt{5}$
c) $D=1$
1d) $C=-2$
e) $ E=\dfrac{13\sqrt{3}}{3}$
Giải thích các bước giải:
b)
$B=(3\sqrt{12}-4\sqrt{3}+\sqrt{15})\sqrt{3}-2\sqrt{5}$
$=(3.2\sqrt{3}-4\sqrt{3}+\sqrt{15})\sqrt{3}-2\sqrt{5}$
$=(2\sqrt{3}+\sqrt{5}.\sqrt{3})\sqrt{3}-2\sqrt{5}$
$=[\sqrt{3}(2+\sqrt{5})]\sqrt{3}-2\sqrt{5}$
$=3(2+\sqrt{5})-2\sqrt{5}$
$=6+3\sqrt{5}-2\sqrt{5}$
$=6+\sqrt{5}$
c)
$D=\dfrac{3}{\sqrt{3}}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-\sqrt{(1-\sqrt{5})^2}$
$=\sqrt{3}+|\sqrt{3}-\sqrt{5}|-|1-\sqrt{5}|$
$=\sqrt{3}+(-\sqrt{3}+\sqrt{5})-(\sqrt{5}-1)$
$=\sqrt{3}-\sqrt{3}+\sqrt{5}-\sqrt{5}+1$
$=1$
d)
$C=\Bigg(\dfrac{\sqrt{15}-\sqrt{20}}{2-\sqrt{3}}+\dfrac{\sqrt{21}-\sqrt{7}}{1-\sqrt{3}}\Bigg):\dfrac{1}{\sqrt{7}-\sqrt{5}}$
$=\Bigg(\dfrac{\sqrt{15}-2\sqrt{5}}{2-\sqrt{3}}-\dfrac{\sqrt{21}-\sqrt{7}}{\sqrt{3}-1}\Bigg)(\sqrt{7}-\sqrt{5})$
$=\Bigg(-\dfrac{2\sqrt{5}-\sqrt{15}}{2-\sqrt{3}}-\dfrac{\sqrt{7}(\sqrt{3}-1)}{\sqrt{3}-1}\Bigg)(\sqrt{7}-\sqrt{5})$
$=\Bigg(-\dfrac{\sqrt{5}(2-\sqrt{3})}{2-\sqrt{3}}-\sqrt{7}\Bigg)(\sqrt{7}-\sqrt{5})$
$=(-\sqrt{5}-\sqrt{7})(\sqrt{7}-\sqrt{5})$
$=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})$
$=-(7-5)$
$=-2$
e)
$E=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}$
$=\dfrac{1}{2}.4\sqrt{3}-\sqrt{\dfrac{33}{11}}+5\dfrac{2}{\sqrt{3}}$
$=2\sqrt{3}-\sqrt{3}+\dfrac{10}{\sqrt{3}}$
$=\sqrt{3}+\dfrac{10}{\sqrt{3}}$
$=\dfrac{13}{\sqrt{3}}$
$=\dfrac{13\sqrt{3}}{3}$
