Question

Найдите значение выражения при к=2-28√2, y=2-8√2

Answer 1

$\frac{k^{2}}{7ky+2k^{2}} : \frac{k}{49 y^{2} -4 k^{2} } = \frac{k^{2}(7y-2k)(7y+2k)}{k*k(7y+2k)} = 7y-2k\\ k=2-28 \sqrt{2}\\ y=2-8 \sqrt{2} \\ 7y-2k=7(2-8 \sqrt{2} )-2(2-28 \sqrt{2} )=\\=14-56 \sqrt{2} -4+56 \sqrt{2} =14-4=10$