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K

khaiproqn81

$\dfrac{1}{\sqrt{3}-\sqrt{2}+1}\\=\dfrac{1+\sqrt{2}+\sqrt{3}}{2+2\sqrt{3}}\\=\dfrac{(2\sqrt{3}-2)(1+\sqrt{2}+\sqrt{3})}{8}\\ =\dfrac{2\sqrt{6}-2\sqrt{2}+4}{8}\\=\dfrac{\sqrt{6}-\sqrt{2}+2}{4}$
 
C

chaugiang81

a.
$\dfrac{1}{\sqrt{3} - \sqrt{2} +1 }$
$= \dfrac{1.(\sqrt{3} + \sqrt{2} -1)}{(\sqrt{3} - \sqrt{2} +1)(\sqrt{3} + \sqrt{2} -1)}$
$= \dfrac{\sqrt{3} + \sqrt{2} -1}{ (\sqrt{3})^2 - (\sqrt{2} -1)^2 }$
$= \dfrac{\sqrt{3} + \sqrt{2} -1}{ 3-2 + 2\sqrt{2} -1}$
$= \dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}}$
$= \dfrac{\sqrt{6} + \sqrt{4} - \sqrt{2} }{4}$
b.
$\dfrac{1}{\sqrt{5} - \sqrt{3} +2 }$
$= \dfrac{\sqrt{5} + \sqrt{3} -2}{(\sqrt{5} - \sqrt{3} +2)(\sqrt{5} +\sqrt{3} -2)}$
$= \dfrac{\sqrt{5} + \sqrt{3} -2}{(\sqrt{5})^2 - (\sqrt{3} -2)^2 }$
$= \dfrac{\sqrt{5} + \sqrt{3} -2}{5- 3+ 4\sqrt{3} -4}$
$= \dfrac{\sqrt{5} + \sqrt{3} -2}{4\sqrt{3} -2}$
$= \dfrac{(\sqrt{5} + \sqrt{3} -2) (4\sqrt{3} +2)}{4\sqrt{3} -2) (4\sqrt{3} +2)}$
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