Question

Решите пожалуйста,,,,,

Answer 1

А)

${x}^{2} - 35 = x \\ {x}^{2} - x - 35 = 0 \\a = 1\\ b = - 1 \\ c = - 35 \\ D = {b}^{2} - 4ac = ( - 1) {}^{2} - 4 \times 1 \times ( - 35) = 1 + 140 = 141 \\ x_{1} = \frac{1 - \sqrt{141} }{2} \\ x_{2} = \frac{1 + \sqrt{141} }{2}$

Б)

${x}^{2} - 4 = 2x - 1 \\ {x}^{2} - 4 - 2x + 1 = 0 \\ {x}^{2} - 2x - 3 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1}  +  x_{2} =2  \\ x_{1} x_{2} = - 3  \\ x_{1} =3  \\ x_{2} = - 1$

В)

$4x - 1 + 5 {x}^{2} = 0 \\ 5 {x}^{2} + 4x - 1 = 0 \\ a =5 \\ b =4 \\ c = - 1\\ D = {b}^{2} - 4ac = {4}^{2} - 4 \times 5 \times ( - 1) = 16 + 20 = 36 \\ x_{1} = \frac{ - 4 - 6}{2 \times 5} = - \frac{10}{10} = - 1 \\ x_{2} = \frac{ - 4 + 6}{2 \times 5} = \frac{2}{10} = 0.2$

Г)

$(x + 2) {}^{2} = (1 - x) {}^{2} \\ (x + 2) {}^{2} - (1 - x) {}^{2} = 0 \\ (x + 2 + 1 - x)(x + 2 - 1 + x) = 0 \\ (0 + 3)(2x + 1) = 0 \\ 2x + 1 = 0 \\ 2x = - 1 \\ x = - 1 \div 2 \\ x = - 0.5$

Д)

$10x + {x}^{2} + 24 = 0 \\ {x}^{2} + 10x + 24 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ x_{1}  +  x_{2} =  - 10\\ x_{1} x_{2} = 24 \\ x_{1} = - 6  \\ x_{2} = - 4$