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simplificand expresia (4x-5)^2-9 supra (4x-3)^2-25 cu 8x-16, x∈R\{-1/2;2} obtinem...

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[tex] \frac{ {(4x - 5)}^{2} - 9 }{ {(4x - 3)}^{2} - 25 } = \frac{ {(4x - 5)}^{2} - {3}^{2} }{ {(4x - 3)}^{2} - 25 } = \frac{(4x - 5 + 3)(4x - 5 - 3)}{ {(4x - 3)}^{2} - 25 } = \frac{(4x - 2)(4x - 5 - 3)}{ {(4x - 3)}^{2} - 25 } = \frac{2(2x - 1)(4x - 5 - 3)}{ {(4x - 3)}^{2} - 25 } = \frac{2(2x - 1)(4x - 8)}{ {(4x - 3)}^{2} - 25 } = \frac{2(2x - 1) \times 4(x - 2)}{ {(4x - 3)}^{2} - 25 } = \frac{8(2x - 1)(x - 2)}{ {(4x - 3)}^{2} - 25 } = \frac{8(2x - 1)(x - 2)}{ {(4x - 3)}^{2} - {5}^{2} } = \frac{8(2x - 1)(x - 2)}{(4x - 3 + 5)(4x - 3 - 5)} = \frac{8(2x - 1)(x - 2)}{(4x + 2)(4x - 3 - 5)} = \frac{8(2x - 1)(x - 2)}{2(2x + 1)(4x - 3 - 5)} = \frac{8(2x - 1)(x - 2)}{2(2x + 1)(4x - 8)} = \frac{8(2x - 1)(x - 2)}{2(2x + 1) \times 4(x - 2)} = \frac{8(2x - 1)(x - 2)}{8(2x + 1)(x - 2)} = \frac{2x - 1}{2x + 1} [/tex]