Вопрос задан 08.06.2023 в 10:16. Предмет Алгебра. Спрашивает Нужненко Макар.

Срочно нужно решить 1)x^2-40=3x 6)3x^2+36=21x2)4x^2=28x-49 7)25+x^2+10x=03)3x^2+5-16x=0

8)2x^2=x+214)10+4x^2-3x=0 9)3x^2-8x+4=05)2x^2-6x=0 10)8+6x^2-x=011) 3x-x^2=0 4-36a^2=0
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Отвечает Герасимчук Машка.

1)x^2-40=3x 6)3x^2+36=21x

2)4x^2=28x-49 7)25+x^2+10x=0

3)3x^2+5-16x=0 8)2x^2=x+21

4)10+4x^2-3x=0 9)3x^2-8x+4=0

5)2x^2-6x=0 10)8+6x^2-x=0

11) 3x-x^2=0

4-36a^2=0

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Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.
  1. To solve the equation x^2 - 40 = 3x, we can rearrange it to x^2 - 3x - 40 = 0. Then, we can factorize the quadratic equation as (x - 8)(x + 5) = 0. Setting each factor equal to zero, we get x - 8 = 0 or x + 5 = 0. Solving for x in each case, we find x = 8 or x = -5.

  2. The equation 4x^2 = 28x - 49 can be rewritten as 4x^2 - 28x + 49 = 0. This equation can be factored as (2x - 7)(2x - 7) = 0, which gives us the repeated root 2x - 7 = 0. Solving for x, we have 2x = 7, and therefore x = 7/2.

  3. To solve the equation 3x^2 + 5 - 16x = 0, we can rearrange it as 3x^2 - 16x + 5 = 0. This equation can be factored as (3x - 1)(x - 5) = 0. Setting each factor equal to zero, we get 3x - 1 = 0 or x - 5 = 0. Solving for x in each case, we find x = 1/3 or x = 5.

  4. The equation 2x^2 = x + 21 can be rearranged as 2x^2 - x - 21 = 0. This equation can be factored as (2x + 7)(x - 3) = 0. Setting each factor equal to zero, we get 2x + 7 = 0 or x - 3 = 0. Solving for x in each case, we find x = -7/2 or x = 3.

  5. To solve the equation 2x^2 - 6x = 0, we can factor out x, giving us x(2x - 6) = 0. Setting each factor equal to zero, we get x = 0 or 2x - 6 = 0. Solving for x in the second case, we find x = 3.

  6. The equation 3x^2 + 36 = 21x can be rearranged as 3x^2 - 21x + 36 = 0. This equation can be factored as (x - 3)(3x - 12) = 0. Setting each factor equal to zero, we get x - 3 = 0 or 3x - 12 = 0. Solving for x in each case, we find x = 3 or x = 4.

  7. The equation 25 + x^2 + 10x = 0 can be rearranged as x^2 + 10x + 25 = 0. This equation can be factored as (x + 5)(x + 5) = 0, which gives us the repeated root x + 5 = 0. Solving for x, we have x = -5.

  8. The equation 2x^2 = x + 21 can be rearranged as 2x^2 - x - 21 = 0. This equation can be factored as (2x + 7)(x - 3) = 0. Setting each factor equal to zero, we get 2x +

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