[tex]2 \sqrt{3x - {x}^{2} - 2 } = \sqrt{x - 1} + \sqrt{2 - x} - 4 + 2x[/tex]

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вопрос

To solve the equation, we can first square both sides to eliminate the square roots:

[4(3x - x^2 - 2) = (x - 1) + 2\sqrt{(x-1)(2-x)} + (2-x) - 8 + 4x]

[12x - 4x^2 - 8 = x - 1 + 2\sqrt{2x-x^2-2} - 8 + 4x - x - 2 + 4x]

[12x - 4x^2 - 8 = 7x - 11 + 2\sqrt{2x-x^2-2}]

Bring all terms to one side of the equation:

[4x^2 - 5x + 3 - 2\sqrt{2x - x^2 - 2} = 0]

Factor the equation if possible:

[(2x - 3)(2x - 1) - 2\sqrt{-(x-2)(x-1)} = 0]

Since the square root term is negative, there is no real solution to this equation.