Question

Помогите пожалуйста!!! Алгебра​

Answer 1

$(3\sqrt{10} + 1) \cdot \left (\dfrac{\sqrt{5}}{\sqrt{8} + \sqrt{5}} - \dfrac{\sqrt{2}}{\sqrt{5} - \sqrt{8}}\right ) = (3\sqrt{10} + 1) \cdot \left(\dfrac{\sqrt{5}}{2\sqrt{2} + \sqrt{5}} - \dfrac{\sqrt{2}}{\sqrt{5} - 2\sqrt{2}}\right ) =\\\\\\= (3\sqrt{10} + 1) \cdot \left(\dfrac{\sqrt{5}(2\sqrt{2} - \sqrt{5})}{(2\sqrt{2} + \sqrt{5})(2\sqrt{2} - \sqrt{5})} - \dfrac{\sqrt{2}(\sqrt{5} + 2\sqrt{2})}{(\sqrt{5} - 2\sqrt{2})(\sqrt{5} + 2\sqrt{2})}\right ) =$

$=(3\sqrt{10} + 1)\cdot \left(\dfrac{2\sqrt{2}\cdot \sqrt{5} - (\sqrt{5})^2}{(2\sqrt{2})^2 - (\sqrt{5})^2} - \dfrac{\sqrt{5}\cdot \sqrt{2} + 2(\sqrt{2})^2}{(\sqrt{5})^2 - (2\sqrt{2})^2}\right ) =\\\\\\=(3\sqrt{10} + 1)\cdot \left (\dfrac{2\sqrt{10}-5}{3} - \dfrac{\sqrt{10}+4}{-3}\right ) = (3\sqrt{10}+1)\cdot \left (\dfrac{2\sqrt{10}-5}{3} + \dfrac{\sqrt{10}+4}{3} \right ) =$

$= (3\sqrt{10} +1)\cdot \dfrac{2\sqrt{10} - 5 + \sqrt{10} + 4}{3} = \dfrac{(3\sqrt{10} + 1)(3\sqrt{10} - 1)}{3} = \dfrac{(3\sqrt{10})^2 - 1}{3} =\\\\\\= \dfrac{90-1}{3} = \dfrac{89}{3} = \boxed{29\dfrac{2}{3}}$