Question

2(x-2)(x+2) = (x-4)(x+4)+(x-3)(x+3)+x Решите уравнение.

Answer 1

2(х-2)(х+2)=(х-4)(х+4)+(х-3)(х+3)+х

Answer 2

2(х-2)(х+2)=(х-4)(х+4)+(х-3)(х+3)+х

2(х^2-4)=х^2-16+х^2-9+х

2х^2-8=2х^2-25+х

2х^2 соответственно сокращаем, получается:

-х=-25+8

-х=-17

х=17

Ответ: 17

Answer 3

To solve the equation, let's first simplify both sides:

Expanding the left-hand side (LHS):

2(x - 2)(x + 2) = 2(x^2 + 2x - 2x - 4) = 2(x^2 - 4)

Expanding the right-hand side (RHS):

(x - 4)(x + 4) + (x - 3)(x + 3) + x = (x^2 + 4x - 4x - 16) + (x^2 + 3x - 3x - 9) + x = x^2 - 25

Now, the equation becomes:

2(x^2 - 4) = x^2 - 25

Next, distribute the 2 on the left-hand side:

2x^2 - 8 = x^2 - 25

Move all terms to one side to set the equation to 0:

2x^2 - x^2 - 8 + 25 = 0

Simplify the equation further:

x^2 + 17 = 0

Now, to solve for x, we'll use the quadratic formula, which is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 0, and c = 17. Plugging these values into the quadratic formula:

x = (± √(0^2 - 4 * 1 * 17)) / 2 * 1 x = (± √(-68)) / 2 x = (± 2i√17) / 2

Now, simplify further:

x = ± i√17

So the two solutions to the equation are:

x = i√17 (where i is the imaginary unit, i.e., √(-1)) x = -i√17

Note: There are no real solutions to the given equation, only imaginary solutions.