Question

X^2/(2x+3)^2-3x/2x+3+2=0

Answer 1

Я решение закрепила)))))

Answer 2

To solve the equation:

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x^2/(2x+3)^2 - 3x/(2x+3) + 2 = 0

Let's simplify the equation step by step:

Step 1: Multiply through by (2x+3)^2 to eliminate the denominators:

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x^2 - 3x(2x+3) + 2(2x+3)^2 = 0

Step 2: Expand the squared terms:

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x^2 - 6x^2 - 9x + 8x^2 + 12x + 18 = 0

Step 3: Combine like terms:

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3x^2 + 3x + 18 = 0

Step 4: Divide the equation by 3 to simplify:

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x^2 + x + 6 = 0

At this point, we have a quadratic equation in standard form. We can solve it using the quadratic formula:

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x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 1, and c = 6. Plugging these values into the quadratic formula:

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x = (-1 ± √(1^2 - 4*1*6)) / (2*1)

Simplifying further:

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x = (-1 ± √(1 - 24)) / 2
x = (-1 ± √(-23)) / 2

Since we have a negative value inside the square root, this equation does not have real solutions. The solutions are complex numbers.

Therefore, the equation x^2/(2x+3)^2 - 3x/(2x+3) + 2 = 0 does not have real solutions.