Question

Помогите, пожалуйста, решить два уравнения

Answer 1

1.

$\frac{4y + 24}{ {5y}^{ 2} - 45 } + \frac{y + 3}{ {5y}^{2} - 15y} = \frac{y - 3}{ {y}^{2} + 3y } \\ \\ \frac{4y + 24}{ 5({y}^{2} - 9) } + \frac{y + 3}{ 5y(y- 3)} - \frac{y - 3}{ y(y+ 3)} =0 \\ \\ \frac{4y + 24}{ 5(y - 3)(y + 3) } + \frac{y + 3}{ 5y(y- 3)} - \frac{y - 3}{ y(y+ 3)} = 0 \\ \\ \frac{y(4y + 24) + {(y + 3)}^{2} - 5 {( y - 3)}^{2} }{5y(y - 3)(y + 3)} =0 \\ \\ \frac{ {4y}^{2} + 24y + {y}^{2} + 6y + 9 - {5y}^{2} + 30y - 45 }{5y(y - 3)(y + 3)} = 0\\ \\ \frac{60y - 36}{5y(y - 3)(y + 3)} = 0 \\ \\ odz \\5y≠0 = >y ≠0 \\ y - 3≠0 = > y ≠3\\ y + 3≠0 \: = > y≠ - 3 \\ \\ 60y - 36 = 0 \\ 60y = 36 \\ y = \frac{36}{60} = \frac{3}{5} = 0.6$

2.

$\frac{ y + 2}{ {8y}^{3} + 1 } - \frac{1}{4y + 2} = \frac{y + 3}{ {8y}^{2} - 4y + 2} \\ \\ \frac{ y + 2}{(2y + 1)({8y}^{2} - 4y+ 2) } - \frac{1}{2(2y + 1)} - \frac{y + 3}{ {8y}^{2} - 4y + 2} = 0 \\ \\ \frac{2(y + 2) - ( {4y}^{2} - 2y + 1) - (2y + 1)(y + 3) }{2(2y + 1)( {4y}^{2} - 2y + 1)} = 0 \\ \\ \frac{2y + 4 - {4y}^{2} + 2y - 1 - {2y}^{2} - 7y - 3}{2(2y + 1)( {4y}^{2} - 2y + 1)} = 0 \\ \\ \frac{ - 3y - {6y}^{2} }{2(2y + 1)( {4y}^{2} - 2y + 1)} = 0 \\ \\ odz \\ 2y + 1≠0 = > y≠ - \frac{1}{2} \\ {4y}^{2} - 2y + 1 ≠0 = > y \notin \: R \\ \\ \frac{ - 3y(1 + 2y)}{2(2y + 1)( {4y}^{2} - 2y + 1) } = 0 \\ \\ \frac{ - 3y}{2( {4y}^{2} - 2y + 1) } = 0 \\ - 3y = 0 \\ y = 0$