Question

хееееелп мииии плиииииз​

Answer 1

Решение:

$x^2-3x-\sqrt{5} -3 = 0\\\\x^2 - 3x -(\sqrt{5} +3) = 0\\\\a = 1\\b = -3\\c = -(\sqrt{5} +3)\\\\D = (-3)^2 - 4 * 1 * (-(\sqrt{5} + 3)) = 9 + 4(\sqrt{5} + 3) = 4\sqrt{5} + 21\\\\x_1 = \frac{3-\sqrt{4\sqrt{5} + 21} }{2} = \frac{3-\sqrt{(1+2\sqrt{5})^2 } }{2} = \frac{3 - 1 - 2\sqrt{5} }{2} = \frac{2(1 - \sqrt{5}) }{2} = 1 - \sqrt{5} \\\\x_2 = \frac{3+\sqrt{4\sqrt{5} + 21} }{2} = \frac{3+\sqrt{(1+2\sqrt{5})^2 } }{2} = \frac{3 + 1 + 2\sqrt{5} }{2} = \frac{2(2 + \sqrt{5}) }{2} = 2 + \sqrt{5}$

$\frac{1}{1-\sqrt{5} } + \frac{1}{2+\sqrt{5} } = \frac{1+\sqrt{5} }{(1-\sqrt{5})(1+\sqrt{5})} + \frac{2-\sqrt{5}}{(2-\sqrt{5})(2+\sqrt{5}) } = \frac{1+\sqrt{5}}{1-5} + \frac{2-\sqrt{5} }{4-5} = -\frac{1+\sqrt{5} }{4} - (2-\sqrt{5} )$

$= \frac{-1-\sqrt{5}-8+4\sqrt{5} }{4} = \frac{-9+3\sqrt{5} }{4}$