Ai giúp mình phần b, c , d với
. mình cần hướng dẫn cách làm ạ
b) VT $=(\dfrac{3\sqrt 2+\sqrt 6}{\sqrt{12}+2}-\dfrac{\sqrt{54}}{3}).\dfrac{2}{\sqrt 6}$
$=\left [\dfrac{\sqrt{6}(\sqrt{3}+1)}{2(\sqrt{3}+1)}-\dfrac{3\sqrt 6}{3} \right ].\dfrac{2}{\sqrt{6}}$
$=\dfrac{\sqrt{6}-2\sqrt{6}}{2}.\dfrac{2}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}.\dfrac{2}{\sqrt{6}}=-1=$ VP
c) VT $=(2+\dfrac{a-\sqrt a}{\sqrt a-1})(2-\dfrac{a+\sqrt a}{1+\sqrt a})$
$=\left [2+\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1} \right ].\left [ 2-\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1} \right ]$
$=(2+\sqrt{a})(2-\sqrt{a})=4-a=$ VP
d) VT $=(\dfrac{3+2\sqrt 3}{\sqrt 3+2}+\dfrac{2+\sqrt 2}{\sqrt 2+1}): (1:\dfrac{1}{\sqrt 2+\sqrt 3})$
$=\dfrac{\sqrt 3(\sqrt 3+2)}{\sqrt 3+2}+\dfrac{\sqrt 2(\sqrt 2+1)}{\sqrt 2+1}: (\sqrt 2+\sqrt 3)$
$=\dfrac{\sqrt 3+\sqrt 2}{\sqrt 2+\sqrt 3}=1=$ VP
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