Question

Doar 14,15,17.
Va rog să mă ajutați

Answer 1

14)

Avem: 

4x² - 1 = 2²x² - 1² = (2x-1)(2x+1).

Expresia devine:


[tex]E(x) = \left(\dfrac{1-12x}{(2x-1)(2x+1)}+\dfrac{x+4}{2x+1}\right)\cdot\dfrac{1}{x-3} [/tex]

Numitorul comun,  in paranteza,  este (2x-1)(2x+1) si dupa ce amplificam a doua fractie , obtinem:


$E(x) = \dfrac{1-12x+2x^2-x+8x-4}{(2x-1)(2x+1)}\cdot\dfrac{1}{x-3} = \\\;\\ = \dfrac{2x^2-5x-3}{(2x-1)(2x+1)}\cdot\dfrac{1}{x-3}$

Descompunem in factori: 2x²-5x-3 =(2x+1)(x-3) si expresia devine:

$\dfrac{(2x+1)(x-3)}{(2x-1)(2x+1)}\cdot\dfrac{1}{x-3} =\dfrac{1}{2x-1}$

b) E(x) ∈ N ⇒ 1/(2x-1) ∈ N ⇒ 2x-1= -1, sau 2x-1 = 1.

Se obtine x ∈ {0, 1}