Question

calcule ((2+3)(3-√3))/((2√5)(3+√3)) obs. a matéria eh sobre racionalização.

Answer 1

$\dfrac{(2+3)\big(3-\sqrt{3}\big)}{2\sqrt{5}\,\big(3+\sqrt{3}\big)}\\\\\\ =\dfrac{5\,\big(3-\sqrt{3}\big)}{2\sqrt{5}\,\big(3+\sqrt{3}\big)}\cdot \dfrac{\sqrt{5}\,\big(3-\sqrt{3}\big)}{\sqrt{5}\,\big(3-\sqrt{3}\big)}\\\\\\ =\dfrac{5\sqrt{5}\,\big(3-\sqrt{3}\big)^2}{2\sqrt{5}\cdot \sqrt{5}\,\big(3+\sqrt{3}\big)\,\big(3-\sqrt{3}\big)}\\\\\\ =\dfrac{\diagup\!\!\!\! 5\sqrt{5}\,\big(3-\sqrt{3}\big)^2}{2\cdot \diagup\!\!\!\! 5\cdot \big(3^2-(\sqrt{3})^2\big)}\\\\\\ =\dfrac{\sqrt{5}\,\big(3-\sqrt{3}\big)^2}{2\cdot (9-3)}\\\\\\ =\dfrac{\sqrt{5}\,\big(3-\sqrt{3}\big)^2}{2\cdot 6}\\\\\\ =\boxed{\begin{array}{c}\dfrac{\sqrt{5}\,\big(3-\sqrt{3}\big)^2}{12} \end{array}}$


( caso queira, pode parar por aqui, mas se quiser expandir o quadrado no numerador... )

$=\dfrac{\sqrt{5}\cdot \big(3^2-2\cdot 3\cdot \sqrt{3}+(\sqrt{3})^2\big)}{12}\\\\\\ =\dfrac{\sqrt{5}\,\big(9-6\sqrt{3}+3\big)}{12}\\\\\\ =\dfrac{\sqrt{5}\,\big(12-6\sqrt{3}\big)}{12}\\\\\\ =\dfrac{\sqrt{5}\cdot \diagup\!\!\!\! 6\cdot \big(2-\sqrt{3}\big)}{\diagup\!\!\!\! 6\cdot 2}\\\\\\ =\boxed{\begin{array}{c}\dfrac{\sqrt{5}\,\big(2-\sqrt{3}\big)}{2} \end{array}}$


Bons estudos! :-)