Предмет: Алгебра, автор: tsokova2018

Помогите пожалуйста с решением

Приложения:

Ответы

Автор ответа: alesyaperova

Ответ:

$\\ 1. \\ a)2 {x}^{2} + 7x - 9 = 0 \\ d = 49 + 4 \times 2 \times 9 = 49 + 72 = 121 = {11}^{2} \\ x1 = \frac{ - 7 + 11}{4} = \frac{4}{4} = 1 \\ x2 = \frac{ - 7 - 11}{4} = - \frac{18}{4} = - 4 \frac{2}{4} = - 4 \frac{1}{2} = - 4.5$

б)

$3 {x}^{2} = 18x \\ 3 {x}^{2} - 18x = 0 \\3x(x - 6) = 0 \\ 3x = 0 \\ x = 0 \div 3 \\ x = 0 \\ x - 6 = 0 \\ x = 0 + 6 \\ x = 6$

ответ:х1=0, х2=6

в)

$100 {x}^{2} - 16 = 0 \\ 100 {x}^{2} = 0 + 16 \\ 100 {x}^{2} = 16 \\ {x}^{2} = 16 \div 100 \\ {x}^{2} = \frac{16}{100} \\ x = \sqrt{ \frac{16}{100} } \\ x1 = \frac{4}{10} = 0.4 \\ x2 = - \frac{4}{10} = - 0.4$

г)

$d) = ( - 16) {}^{2} - 4 \times 1 \times 63 = 256 - 252 = 4 = {2}^{2} \\ x1 = \frac{16 + 2}{2} = \frac{18}{2} = 9 \\ x2 = \frac{16 - 2}{2} = \frac{14}{2} = 7$

а)

${x}^{2} - 12 + x = 0 \\ {x}^{2} + x - 12 = 0 \\ d =1 + 48 = 49 = {7}^{2} \\ x1 = \frac{ - 1 + 7}{2} = \frac{6}{2} = 3 \\ x2 = \frac{ - 1 - 7}{2} = - \frac{8}{2} = - 4$

б)

$6x + 5(x - 2) = 3x(x - 2) \\ 6x + 5x - 10 = 3 {x}^{2} - 6x \\ 11x - 10 = 3 {x}^{2} - 6x \\ 11x - 10 - 3 {x}^{2} + 6x = 0 \\ - 3 {x}^{2} + 17x - 10 = 0 \times ( - 1) \\ 3 {x}^{2} - 17x + 10 = 0 \\d = 289 - 120 = 169 = {13}^{2} \\ x1 = \frac{17 + 13}{6} = \frac{30}{6} = 5 \\ x2 = \frac{17 - 13}{6} = \frac{4}{6} = \frac{2}{3}$

в)

$(1 + 3x)(1 + 2x) = (5 - 3x)(1 - 2x) \\ 1 + 2x + 3x + 6 {x}^{2} = 5 - 10x - 3x + 6 {x}^{2} \\ 1 + 5x + 6 {x}^{2} = 5 - 13x + 6 {x}^{2} \\ 1 + 5x + 6 {x}^{2} - 5 + 13x - 6 {x}^{2} = 0 \\ 18x - 4 = 0 \\ 18x = 0 + 4 \\ 18x = 4 \\ x = 4 \div 18 \\ x = \frac{4}{18} = \frac{2}{9}$