(х+3)(2х+1)/х^2-4=0;

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вопрос

To solve this equation, we first need to factor the numerator of the fraction:

(х+3)(2х+1)

Now we have the equation in the form (х+3)(2х+1) / х^2 - 4 = 0

Next, we factor the denominator:

х^2 - 4 = (x + 2)(x - 2)

Now our equation looks like this: (х+3)(2х+1) / (x + 2)(x - 2) = 0

To find the values of x that satisfy the equation, we set each factor in the numerator equal to zero:

х + 3 = 0
x = -3

2х + 1 = 0
2x = -1
x = -1/2

Now we check the values of x that make the denominator equal to zero to make sure they are not solutions:

x + 2 = 0
x = -2

x - 2 = 0
x = 2

Since x = -2 and x = 2 make the denominator equal to zero, they are not valid solutions.

Therefore, the solutions to the equation are x = -3 and x = -1/2.