3x/2x+5-28-53x/4x²-25=4x/2x-5

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вопрос

To solve this equation, we first need to simplify both sides of the equation.

Starting with the left side:

3x/(2x+5) - 28 - 53x/(4x^2-25)
= 3x/(2x+5) - 28 - 53x/[(2x+5)(2x-5)] (Difference of squares: 4x^2 - 25 = (2x+5)(2x-5))
= [3x(2x-5) - 28(2x+5) - 53x]/[(2x+5)(2x-5)] (Common denominator)
= (6x^2 - 15x - 56x - 140 - 53x)/[(2x+5)(2x-5)] (Distribute)
= (6x^2 - 124x - 140)/[(2x+5)(2x-5)] (Combine like terms)

Now moving on to the right side:

4x/(2x-5)
= 2(2x)/(2x-5)
= 4x/(2x-5)

Setting the two sides equal to each other:

(6x^2 - 124x - 140)/[(2x+5)(2x-5)] = 4x/(2x-5)

Now, we can cross multiply to get rid of the denominators:

(6x^2 - 124x - 140) = 4x(2x+5)
6x^2 - 124x - 140 = 8x^2 + 20x

Rearranging the equation:

2x^2 + 144x + 140 = 0

This is now a quadratic equation that can be solved using factoring, completing the square, or the quadratic formula.