Question

ПОЖАЛУЙСТА ПОМОГИТЕ!!!!​

Answer 1

$x\neq0 \: \: \: \: \: x\neq - 3 \\ \frac{5}{x + 3} - \frac{3}{x} = \frac{2 - x}{ {x}^{2} + 3x } \\ \frac{5}{x + 3} - \frac{3}{x} - \frac{2 - x}{x(x + 3)} = 0 \\ \frac{5x - 3(x + 3) - (2 - x)}{x(x + 3)} = 0 \\ \frac{5x - 3x - 9 - 2 + x}{x(x + 3)} = 0 \\ \frac{3x - 11}{x(x + 3)} = 0 \\ 3x - 11 = 0 \\ 3x = 11 \\ x = \frac{11}{3} \\ x = 3 \frac{2}{3}$

${x}^{4} - 6 {x}^{2} + 5 = 0 \\ {x}^{2} = a \: , \: \: a \geqslant 0 \\ {a}^{2} - 6a + 5 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ a_{1} + a_{2} = 6 \\ a_{1}a_{2} = 5 \\ a_{1} = 1 \\ a _{2}= 5 \\ \\ {x}^{2} = 1 \\ {x}^{2} = 5 \\ \\ x_{1} = - 1 \\ x_{2} = 1\\ x_{3} = - \sqrt{5} \\ x_{4} = \sqrt{5}$