Question

a)$\frac{x+1}{x+2} * \frac{x+3}{x+4} : \frac{ x^{2} +6x+5}{ x^{2} +6x+8} : \frac{x+3}{x+5}$
b) $( \frac{2}{x+2} + \frac{x+3}{ x^{2} -4} - \frac{3x+1}{ x^{2} -4x+4} ): \frac{14}{4- x^{2} }$

Answer 1

    
$a)\;\; \frac{x+1}{x+2} * \frac{x+3}{x+4} : \frac{ x^{2} +6x+5}{ x^{2} +6x+8} : \frac{x+3}{x+5} = \\ \\ =\frac{x+1}{x+2} * \frac{x+3}{x+4} * \frac{ x^{2} +6x+8}{ x^{2} +6x+5} * \frac{x+5}{x+3} = \\ \\ =\frac{x+1}{x+2} * \frac{x+3}{x+4} * \frac{ (x+2)(x+4)}{(x+1)(x+5)} * \frac{x+5}{x+3} = \\ \\ = \frac{(x+1)(x+3)(x+2)(x+4)(x+5)}{(x+2)(x+4)(x+1)(x+5)(x+3)}=1$



$b)\;( \frac{2}{x+2} + \frac{x+3}{ x^{2} -4} - \frac{3x+1}{ x^{2} -4x+4} ): \frac{14}{4- x^{2}}= \\ \\ =( \frac{2}{x+2} + \frac{x+3}{ (x-2)(x+2)} - \frac{3x+1}{ (x-2)^{2}} )* \frac{(2-x)(2+x)}{14}= \\ \\ \text{Factorul comun este: }(x-2)^{2}(x+2) \\ \\ = \frac{2(x-2)^{2}+(x+3)(x-2)-(3x+1)(x+2)}{(x-2)^{2}(x+2)}* \frac{(2-x)(2+x)}{14}= \\ \\ = \frac{2x^{2}-8x+8+x^{2}+x-6-3x^{2}-7x-2)}{(x-2)}* \frac{-1}{14}= \\ \\ = \frac{-14x}{x-2}* \frac{-1}{14}= \frac{x}{x-2}$