Question

Ex 4!!! Repedee dau 100 de puncte si coroana !! Aveti poza atașata!!!

Answer 1

a)

$\dfrac{-2}{x-2} + \dfrac{x}{x+2} - \dfrac{1}{x^{2} -4x + 4} = -\dfrac{^{(x-2)(x+2)} 2}{x-2} + \dfrac{^{(x-2)^{2})}x}{x+2} - \dfrac{^{x+2)} 1}{(x-2)^{2}} =$

$= \dfrac{-2(x-2)(x+2) + x(x-2)^{2} - x -2}{(x+2)(x-2)^{2}} = \dfrac{-2(x^{2} -4) + x(x^{2}-4x+4) - x-2}{(x+2)(x-2)^{2}}$

$= \dfrac{-2x^{2} + 8 + x^{3} - 4x^{2} + 4x - x - 2}{(x+2)(x-2)^{2}} = \dfrac{x^{3} - 6x^{2} + 3x + 6}{(x+2)(x-2)^{2}}$

$= \dfrac{x^{3} - 2x^{2} - 4x^{2} + 8x - 5x + 10}{(x+2)(x-2)^{2}} = \dfrac{x^{2}(x - 2) - 4x(x - 2) - 5(x - 2)}{(x+2)(x-2)^{2}}$

$= \dfrac{(x - 2)(x^{2} - 4x - 5)}{(x+2)(x-2)^{2}} = \dfrac{x(x - 5) + (x - 5)}{(x+2)(x-2)} = \dfrac{(x + 1)(x - 5)}{(x+2)(x-2)}$

b)

$\dfrac{x^{2} + 2}{2x} - \dfrac{x^{2}}{2x + 4} + \dfrac{4}{x^{2} + 2x} = \dfrac{^{x + 2)} x^{2} + 2}{2x} - \dfrac{^{x)} x^{2}}{2(x + 2)} + \dfrac{^{2)}4}{x(x + 2)} =$

$= \dfrac{(x^{2} + 2)(x + 2) - x^{3} + 8}{2x(x + 2)} = \dfrac{x^{3} + 2x^{2} + 2x + 4 - x^{3} + 8}{2x(x + 2)}$

$= \dfrac{2(x^{2} + x + 6)}{2x(x + 2)} = \dfrac{x^{2} + x + 6}{x(x + 2)}$

c)

$\dfrac{x}{2 - x} - \dfrac{2 - x}{x} + \dfrac{4x - 4}{(x - 1)^{2} - 1} = -\dfrac{x}{x - 2} + \dfrac{x - 2}{x} + \dfrac{4x - 4}{x^{2} - 2x + 1 - 1}$

$= -\dfrac{^{x)} x}{x - 2} + \dfrac{^{x - 2)}x - 2}{x} + \dfrac{4x - 4}{x(x - 2)} = \dfrac{-x^{2} + (x - 2)^{2} + 4x - 4}{x(x - 2)}$

$= \dfrac{-x^{2} + x^{2} -4x + 4 + 4x - 4}{x(x - 2)} = \dfrac{0}{x(x - 2)} = 0$