Question

Избавиться от иррацианальности в знаменателе дроби
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Answer 1

$\dfrac{3}{\sqrt5} = \dfrac{3}{\sqrt5} * \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{3\sqrt{5}}{\sqrt{5}\sqrt{5}} = \dfrac{3\sqrt{5}}{5}$

$\dfrac{5}{2\sqrt3} = \dfrac{\sqrt3}{\sqrt3} = \dfrac{5\sqrt3}{2\sqrt3\sqrt3} = \dfrac{5\sqrt3}{2*3} = \dfrac{5\sqrt3}{6}$

$\dfrac{10}{\sqrt{10}-3} = \dfrac{10}{\sqrt{10}-3} *\dfrac{\sqrt{21}+3}{\sqrt{21}+3} = \dfrac{10(\sqrt{21}+3)}{(\sqrt{21}-3)(\sqrt{21}+3)} = \dfrac{10(\sqrt{21}+3)}{21-9} =\\\\= \dfrac{10(\sqrt{21}+3)}{12} = \dfrac{5(\sqrt{21}+3)}{6} = \dfrac{5\sqrt{21}+15}{6}$

Answer 2

а) 3/√5 = (3√5)/(√5*√5) = (3√5)/5

б) 5/(2√3) = (5*2√3)/(2√3 * 2√3) = (10√3)/12 = (5√3)/6

в) 10/(√21-3) = 10(√21+3)/(√21-3)(√21+3) = 10(√21+3)/(21-9) = 5(√21+3)/6