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1) \( \frac{3 x+1}{x+2}-1=\frac{x-1}{x-2} \)

Ask by Sanders Gibson. in Kazakhstan
Jan 19,2025

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Решение уравнения: \[ \frac{3x + 1}{x + 2} - 1 = \frac{x - 1}{x - 2} \] Получаем: \[ x = 3 + \sqrt{5} \quad \text{или} \quad x = 3 - \sqrt{5} \]

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To solve the equation \( \frac{3x+1}{x+2} - 1 = \frac{x-1}{x-2} \), let's first simplify the left side. Starting with \( \frac{3x+1}{x+2} - 1 \): \[ \frac{3x+1}{x+2} - \frac{x+2}{x+2} = \frac{(3x+1) - (x+2)}{x+2} = \frac{2x - 1}{x+2} \] Now we can rewrite the original equation as: \[ \frac{2x - 1}{x + 2} = \frac{x - 1}{x - 2} \] Next, we can cross-multiply to eliminate the fractions: \[ (2x - 1)(x - 2) = (x - 1)(x + 2) \] Expanding both sides: Left Side: \[ 2x^2 - 4x - x + 2 = 2x^2 - 5x + 2 \] Right Side: \[ x^2 + 2x - x + 2 = x^2 + x + 2 \] Setting both sides equal gives us: \[ 2x^2 - 5x + 2 = x^2 + x + 2 \] Now, subtract \( x^2 + x + 2 \) from both sides: \[ 2x^2 - 5x + 2 - x^2 - x - 2 = 0 \] This simplifies to: \[ x^2 - 6x = 0 \] Factoring out \( x \): \[ x(x - 6) = 0 \] Setting each factor to zero gives the solutions: \[ x = 0 \quad \text{or} \quad x = 6 \] Thus, the solutions to the equation are: \[ x = 0 \quad \text{and} \quad x = 6 \]

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