Question

como racionaliza isso aqui ? :o

Answer 1

Para racionalizar o denominador, multiplica-se o numerador e o denominador pelo conjugado do denominador, que é  $2\sqrt{2}-\sqrt{5}$:

$\dfrac{9\sqrt{2}}{2\sqrt{2}+\sqrt{5}}\\ \\ =\dfrac{9\sqrt{2}\cdot \left(2\sqrt{2}-\sqrt{5} \right )}{\left(2\sqrt{2}+\sqrt{5} \right )\cdot \left(2\sqrt{2}-\sqrt{5} \right )}\\ \\ =\dfrac{9\sqrt{2}\cdot 2\sqrt{2}-9\sqrt{2}\cdot \sqrt{5}}{\left(2\sqrt{2} \right )^{2}-\left(\sqrt{5} \right )^{2}}\\ \\ =\dfrac{18\cdot \sqrt{2\cdot 2}-9\sqrt{2\cdot 5}}{2^{2}\cdot \sqrt{2^{2}}-\sqrt{5^{2}}}\\ \\ =\dfrac{18\cdot \sqrt{2^{2}}-9\sqrt{10}}{4\cdot 2-5}\\ \\ =\dfrac{18\cdot 2-9\sqrt{10}}{8-5}\\ \\ =\dfrac{36-9\sqrt{10}}{8-5}\\ \\ =\dfrac{\diagup\!\!\!\! 3\cdot \left(12-3\sqrt{10} \right )}{\diagup\!\!\!\! 3}\\ \\ =12-3\sqrt{10}$

Answer 2

Oie, Td Bom?!

$= \frac{9 \sqrt{2} }{2 \sqrt{2} + \sqrt{5} }$

$= \frac{9 \sqrt{2} }{2 \sqrt{2} + \sqrt{5} } \times \frac{2 \sqrt{2} - \sqrt{5} }{2 \sqrt{2} - \sqrt{5} }$

$= \frac{9 \sqrt{2} \times (2 \sqrt{2} - 5)}{(2 \sqrt{2} + \sqrt{5} ) \times (2 \sqrt{2} - 5) }$

$= \frac{9 \sqrt{2} \times (2 \sqrt{2} - \sqrt{5}) }{8 - 5}$

$= \frac{9 \sqrt{2} \times (2 \sqrt{2} - 5)}{3}$

$= 3 \sqrt{5} \times (2 \sqrt{2} - \sqrt{5} )$

$= 3 \sqrt{2} \times 2 \sqrt{2} - 3 \sqrt{2} \sqrt{5}$

$= 12 - 3 \sqrt{10}$

Att. Makaveli1996