Question

Умоляю
Должен получиться 1

Answer 1

$\frac{(2y - 1) {( {(2y + 1)}^{2} - 8y) }^{ - \frac{1}{2} } }{25 {y}^{2} - 3} = \\ = \frac{2y - 1}{ {(4 {y}^{2} + 4y + 1 - 8y) }^{ \frac{1}{2} } (25 {y}^{2} - 3) } = \\ = \frac{2y - 1}{ {(4 {y}^{2} - 4y + 1)}^{ \frac{1}{2} }(25 {y}^{2} - 3) } = \\ = \frac{2y - 1}{ \sqrt{ {(2y - 1)}^{2} }(25 {y}^{2} - 3) } = \\ = \frac{2y - 1}{ |2y - 1|(25 {y}^{2} - 3) }$

При у=√2/5

$2y - 1 = \frac{2 \sqrt{2} }{5} - 1 = \\ = \frac{2 \sqrt{2} - 5 }{5} = \frac{(2 \sqrt{2} - 5)(2 \sqrt{2} + 5) }{5(2 \sqrt{2} + 5) } = \\ = \frac{ {(2 \sqrt{2} )}^{2} - {5}^{2} }{5(2 \sqrt{2} + 5) } = \frac{8 - 25}{5(2 \sqrt{2} + 5) } = \\ = \frac{ -17 }{5(2 \sqrt{2} + 5)} < 0$

а значит

$|2y - 1| = - (2y - 1)$

Таким образом,

$\frac{2y - 1}{ |2y - 1| (25 {y}^{2} - 3) } = \\ = \frac{2y - 1}{ - (2y - 1)(25 {y}^{2} - 3) } = \\ = - \frac{1}{25 {y}^{2} - 3}$

Подставляя у=√2/5 получаем:

$- \frac{1}{25 {( \frac{ \sqrt{2} }{5}) }^{2} - 3 } = - \frac{1}{25 \times \frac{2}{25} - 3 } = \\ = - \frac{1}{2 - 3} = 1$