Question

срочно допоможіть. дам 35б​

Answer 1

$( {x}^{2} - 6x + 1) {}^{2} + 6( {x}^{2} - 6x + 1) - 7 = 0 \\ {x}^{2} - 6x + 1 = a \\ {a}^{2} + 6a - 7 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ a _{1}+ a_{2} = - 6 \\ a_{1}a_{2} = - 7 \\ a_{1} = - 7 \\ a _{2}= 1 \\ \\ 1) \: a = - 7 \\ {x}^{2} - 6x + 1 = - 7 \\ {x}^{2} - 6x + 8 = 0 \\ po \: \: \: teorme \: \: \: vieta \\ x_{1} + x_{2} = 6 \\ x_{1} x_{2} =8 \\ x_{1} = 2\\ x_{2} = 4 \\ \\ 2) \: a = 1 \\ {x}^{2} - 6x + 1 = 1 \\ {x}^{2} - 6x = 0 \\ x(x - 6) = 0 \\ x_{3} = 0 \\ x_{4} = 6$