Question

Задача № 1 : Решите уравнение (x-2)(x-3)(x+4)(x+5) = 1320. срочно!!!!!!!!!!плиз

Answer 1

Перемножаем первую и третью, вторую и четвертую скобки:
(x^2 + 2x - 8)(x^2 + 2x - 15) = 1320
Замена: x^2 + 2x - 15 = y;
(y + 7)y = 1320
y^2 + 7y - 1320 = 0
Тут корни проще угадать. Заметим, что 1320 = 33 * 40, тогда решение этого уравнения y = 33 и y = -40.

x^2 + 2x - 15 = 33
x^2 + 2x + 1 = 49
x = -1 +- 7
x = -8 или x = 6

x^2 + 2x - 8 = -40
x^2 + 2x + 1 = -31
корней нет


Ответ. -8, 6.

Answer 2

Task 1: Solving the equation (x-2)(x-3)(x+4)(x+5) = 1320

To solve the equation (x-2)(x-3)(x+4)(x+5) = 1320, we can follow these steps:

Step 1: Expand the equation Step 2: Simplify the equation Step 3: Solve for x

Let's go through each step in detail:

Step 1: Expand the equation To expand the equation, we can use the distributive property. Multiply each term in the first set of parentheses by each term in the second set of parentheses, and then multiply the resulting expression by the third set of parentheses, and finally by the fourth set of parentheses.

(x-2)(x-3)(x+4)(x+5) = 1320

Expanding the equation, we get:

x^4 + 4x^3 - 3x^2 - 12x^3 - 48x^2 + 36x + 4x^2 + 16x - 12 - 15x - 60 + 45 = 1320

Simplifying the equation, we have:

x^4 - 11x^3 - 7x^2 + 61x - 27 = 1320

Step 2: Simplify the equation To simplify the equation, we need to move all the terms to one side of the equation, so that it becomes a quadratic equation equal to zero.

x^4 - 11x^3 - 7x^2 + 61x - 27 - 1320 = 0

x^4 - 11x^3 - 7x^2 + 61x - 1347 = 0

Step 3: Solve for x To solve the equation, we can use numerical methods or factorization. Since the equation is a quartic equation, it may not have simple factorizations. Therefore, we can use numerical methods such as graphing or using a calculator to find the approximate solutions.

Unfortunately, without further information or constraints, it is not possible to determine the exact solutions to this equation.

I hope this helps! Let me know if you have any further questions.