Question

Find the solution of the attached file.
I need urgent​

Answer 1

Answer:

$\dfrac{1}{x} - \dfrac{1}{x + b} = \dfrac{1}{a} - \dfrac{1}{a + b} \\ \\ \dfrac{x + b - x}{x(x + b)} = \dfrac{a + b - a}{a(a + b)} \\ \\ \dfrac{b}{ {x}^{2} + bx} = \dfrac{b}{ {a}^{2} + ab} \\ \\ \frac{1}{ {x}^{2} + bx } = \frac{1}{ {a}^{2} + ab } \\ \\ {a}^{2} + ab = {x}^{2} + bx \\ \\ {a}^{2} + ab - {x}^{2} - bx = 0 \\ \\ {x}^{2} - {a}^{2} + bx - ab = 0 \\ \\ (x + a)(x - a) + b(x - a) = 0 \\ \\ (x - a) \{x + a + b) \}$

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