Question

СРОЧНО АЛГЕБРА дам 40 балов
Объяснение писать обязательно

Answer 1

1)

$2 {x}^{2} - 12 x+ 10 =0 \\ {x}^{2} - 6x + 5 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = 6 \\ x_{1} x_{2} =5 \\ x_{1} = 5\\ x_{2} = 1 \\ \\ a {x}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 2 {x}^{2} - 12x + 10 = 2(x - 5)(x - 1)$

2)

${x}^{3} - 125 = (x - 5)( {x}^{2} + 5x + 25)$

3)

$\frac{2 {x}^{2} - 12x + 10 }{ {x}^{3} - 125} = \frac{2(x - 5)(x - 1)}{(x - 5)( {x}^{2} + 5x + 25)} = \frac{2(x - 1)}{ {x}^{2} + 5x + 25 }$