Question

Calcule:
$\frac{90\sqrt{5}-30 \sqrt{5}}{240+120 \sqrt{3} }$

Answer 1

(90√5) - (30√5) = 30[(3√5) - √5)] = 30(2√5)

240 + 120√3 = 120(2 + √3)

$\frac{30(2 \sqrt{5} )}{120(2+ \sqrt{3} )}$ = $\frac{2 \sqrt{5} }{4(2+ \sqrt{3} )}$

Answer 2

$\dfrac{90\sqrt{5}-30\sqrt{5}}{240+120\sqrt{3}}\\\\\\ =\dfrac{(90-30)\sqrt{5}}{240+120\sqrt{3}}\\\\\\ =\dfrac{60\sqrt{5}}{120\cdot \big(2+\sqrt{3}\big)}\\\\\\ =\dfrac{\diagup\!\!\!\!\! 60\sqrt{5}}{\diagup\!\!\!\!\! 60\cdot 2\cdot \big(2+\sqrt{3}\big)}\\\\\\ =\dfrac{\sqrt{5}}{2\cdot \big(2+\sqrt{3}\big)}$


Multiplicando e dividindo pelo conjugado $\big(2-\sqrt{3}\big):$

$=\dfrac{\sqrt{5}}{2\cdot \big(2+\sqrt{3}\big)}\cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\\\\\\ =\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2\cdot \big(2+\sqrt{3}\big)\big(2-\sqrt{3}\big)}$

$=\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2\cdot \big(2^2-(\sqrt{3})^2\big)}\\\\\\ =\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2\cdot (4-3)}\\\\\\ =\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2\cdot 1}\\\\\\ =\boxed{\begin{array}{c}\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2} \end{array}}$


Bons estudos! :-)