2/x-5-4/x+5=6/x^2-25

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вопрос

To solve this equation, we first need to find a common denominator for the fractions on the left side of the equation:

2/x - 5 - 4/x + 5 = 6/(x^2 - 25)

The common denominator for the fractions 2/x and 4/x is x, so we can rewrite the equation as:

(2 - 5x)/x - 5 = 6/(x^2 - 25)

Now, we need to factor the denominator on the right side of the equation:

x^2 - 25 = (x + 5)(x - 5)

Now, we can rewrite the equation as:

(2 - 5x)/x - 5 = 6/((x + 5)(x - 5))

Next, we can multiply both sides of the equation by x(x + 5)(x - 5) to eliminate the denominators:

2(x + 5) - 5x(x + 5)(x - 5) = 6x

Now, we can expand and simplify the equation:

2x + 10 - 5x^2 - 5x^2 + 25x = 6x
2x + 10 - 10x^2 + 25x = 6x
-10x^2 + 27x + 10 = 6x
-10x^2 + 21x + 10 = 0

Finally, we can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -10, b = 21, and c = 10:

x = (-21 ± √(21^2 - 4(-10)(10))) / 2(-10)
x = (-21 ± √(441 + 400)) / -20
x = (-21 ± √841) / -20
x = (-21 ± 29) / -20

So the solutions for x are:

x = (29 - 21) / -20 = 8 / -20 = -0.4

x = (-29 - 21) / -20 = -50 / -20 = 2.5

Therefore, the solutions to the equation are x = -0.4 and x = 2.5.