Question

50 БАЛЛОВ! Помогите пожалуйста. 11-й класс​

Answer 1

1)

$(x - 1)(x - 7)(x - 4)(x + 2) = 40 \\ (x {}^{2} - 7x - x + 7)( {x}^{2} + 2x - 4x - 8) - 40 = 0 \\ ( {x}^{2} - 8x + 7)( {x}^{2} - 2x - 8) - 40 = 0 \\ {x}^{4} - 2 {x}^{3} - 8x {}^{2} - 8 {x}^{3} + 16 {x}^{2} + 64x + 7{x}^{2} - 14x - 56 - 40= 0 \\ {x}^{4} - 10 {x}^{3} + 15 {x}^{2} + 50x - 96 = 0$

Разложим на множители с помощью схемы Горнера (фото):

$(x -2 )(x {}^{3} - 8 {x}^{2} - x + 48) = 0 \\ (x - 2)(x - 3)( {x}^{2} - 5x - 16) = 0 \\ x _{1 } = 2 \\ x _{2} = 3 \\ {x}^{2} - 5x - 16 = 0 \\ d = ( - 5) {}^{2} - 4 \times 1 \times ( - 16) = \\ 25 + 64 = 89 \\ x _{3} = \frac{5 + \sqrt{89} }{2} \\ x _{4} = \frac{5 - \sqrt{89} }{2}$

2)

$6( {x}^{2} + \frac{1}{ {x}^{2} } )+ 5(x + \frac{1}{x} ) - 38 = 0 \\ 6 {x}^{2} + \frac{6}{ {x}^{2} } + 5x + \frac{5}{x} - 38 = 0 \\ \frac{6 {x}^{4} + 6 + 5 {x}^{3} + 5x - 3 8{x}^{2} }{ {x}^{2} } = 0 \\ 6 {x}^{4} + 5 {x}^{3} - 3 8{x}^{2} + 5x + 6 = 0$

Разложим на множители с помощью схемы Горнера (фото):

$(x - 2)(6 {x}^{3} + 17 {x}^{2} - 4x - 3) = 0 \\ (x - 2)(x + 3 )(6 {x}^{2} - x - 1) = 0 \\ x _{1} = 2 \\x _{2} = - 3 \\ 6 {x}^{2} - x - 1 1 = 0 \\ d = ( - 1) {}^{2} - 4 \times 6 \times( - 1) = \\ 1 + 24 = 25 \\ x _{3} = \frac{1 + 5}{2 \times 6} = \frac{6}{12} = \frac{1}{2} = 0.5 \\ x _{4} = \frac{1 - 5}{2 \times 6} = \frac{4}{12} = \frac{1}{3}$