• Предмет: Алгебра
  • Автор: maxbessmertnui
  • Вопрос задан 8 лет назад

Помогите Пожалуйста!!!
1,2,3 номер​

Приложения:

Ответы

Ответ дал: prlkomn
0

#1

1)(х-8)*(х+4)

 {x}^{2}  - 4x - 32 = 0 \ d = ( - 4) ^{2}  - 4 times 1 times ( - 32) = 16 + 128 = 144 =  {12}^{2}  \ x1 =  frac{4 + 12}{2}  =  frac{16}{2}  = 8 \ x2 =  frac{4 - 12}{2}  =  frac{ - 8}{2}  =  - 4

2)(х-3)*(х-12)

4 {x}^{2}  - 15x + 9 = 0 \ d = ( - 15)^{2}  - 4  times 4 times 9 = 225 - 144 = 81 =  {9}^{2}  \ x1 =  frac{15 - 9}{2}  =  frac{6}{2}  = 3 \ x2 =  frac{15 + 9}{2}  =  frac{24}{2}  = 12

#2

1)

 {x}^{2}  = y \ y^{2}  - 8y- 9 = 0 \ d = ( - 8)^{2}  - 4 times 1 times ( - 9) = 64 + 36 = 100 =  {10}^{2}  \ y1 =  frac{8 - 10}{2}  =  frac{ - 2}{2}  =  - 1 \ y2 =  frac{8 + 10}{2}  =  frac{18}{2}  = 9 \ x1 =  -  \ x2 =  sqrt{9}  =  + - 3

2)

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

#3

 frac{x - 1}{x + 2}  +  frac{x + 1}{x  - 2}   +  frac{2x + 8}{4  -  {x}^{2} }  = 0 \  frac{x - 1}{x + 2}  + frac{x + 1}{x - 2}  +  frac{2x + 8}{(2 - x) times (2 + x)}  = 0 \  frac{ {x}^{2} - x - 2x + 2 +  {x}^{2}   + x + 2x + 2 - 2x - 8}{(2 - x) times (2 + x)}  = 0 \  frac{2 {x}^{2}  - 4 - 2x}{(2 - x) times (2 + x)}  = 0 \  frac{2 times (x + 1) times (x - 2)}{(x - 2) times (x + 2)}  = 0 \  frac{2 times (x + 1)}{x + 2}  = 0 \  frac{2x + 2}{x + 2}  = 0 \ 2x + 2 = 0 \ 2x =  - 2 \ x =  - 1

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