Question

 a) $\frac{x}{ x^{2} -1} - \frac{2x-3}{ x^{2} -2x+1} + \frac{1}{x+1} ): \frac{4- x^{2} }{ x^{3} + x^{2} -x-1}$
b) $( \frac{ x^{2} -x}{ x^{2} -25} - \frac{x-2}{x+5} + \frac{x-3}{5-x} )* \frac{ x^{2} +2x-15}{ x^{2} -x-2}$

Answer 1

a) numitorul comun este ( x + 1)( x - 1)²
   [ x / ( x - 1)( x + 1) - ( 2x - 3) / ( x - 1)² + 1 / ( x + 1)] :
         : ( 4 - x²) / [( x³ + x²) - ( x + 1)] =
= [ x( x - 1) / ( x + 1)( x - 1)² - ( 2x - 3)( x + 1) / ( x + 1)( x - 1)² + ( x - 1)² /
       / ( x + 1)( x - 1)²] : ( 4 - x²) / [ x²( x + 1) - ( x + 1)] =
= [ x ( x - 1) - ( 2x - 3)( x + 1) + ( x - 1)² totul supra ( x + 1)( x - 1)² ] :
       : ( 4 - x²) / ( x + 1)( x² - 1)
= ] x² - x - ( 2x² + 2x - 3x - 3) + x² - 2x + 1 totul supra ( x + 1)( x - 1)²] :
       : ( 4 - x²) / ( x + 1)( x + 1)( x - 1)
= [ x² - x  - ( 2x² - x - 3) + x² - 2x + 1) totul supra ( x + 1)( x - 1)² ] :
       : ( 4 - x²) / ( x + 1)²( x - 1)
= ( 2x² - 3x + 1 - 2x² + x + 3) totul supra ( x + 1)( x - 1)²] :
       : ( 4 - x²) / ( x + 1)²( x - 1)
= ( - 2x + 4) / ( x + 1)( x - 1)²  :  ( 4 - x²) / ( x + 1)²( x - 1)
= ( 4 - 2x) / ( x + 1)( x - 1)² · ( x + 1)²( x - 1) / ( 4 - x²)
se simplifica ( x - 1)² cu ( x - 1)    si ( x + 1)² cu ( x + 1)
= ( 4 - 2x)( x + 1) totul supra ( 4 - x²)( x - 1)
= 2( 2 - x)( x + 1) totul supra ( 2 - x)( 2 + x)( x - 1), se simplifica
= 2( x + 1) / ( 2 + x)( x - 1)  ,  / = supra

Answer 2

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


                     
$b)\;( \frac{ x^{2} -x}{ x^{2} -25} - \frac{x-2}{x+5} + \frac{x-3}{5-x} )* \frac{ x^{2} +2x-15}{ x^{2} -x-2}= \\ \\ =( \frac{ x^{2} -x}{ x^{2} -25} - \frac{(x-2)(x-5)}{x^{2} -25} - \frac{(x-3)(x+5)}{x^{2} -25} )* \frac{ (x+5)(x-3)}{ (x-2)(x+1)}= \\ \\ =\frac{ (x^{2} -x)-(x-2)(x-5)-(x-3)(x+5)}{ x^{2} -25}* \frac{ (x+5)(x-3)}{ (x-2)(x+1)}= \\ \\ =\frac{ x^{2} -x-x^{2}+7x-10-x^{2}-2x+15}{ x^{2} -25}* \frac{ (x+5)(x-3)}{ (x-2)(x+1)}=$

$=\frac{ -x^{2}+4x+5}{ x^{2} -25}* \frac{ (x+5)(x-3)}{ (x-2)(x+1)}= \\ \\=\frac{ -(x-5)(x+1)}{ x^{2} -25}* \frac{ (x+5)(x-3)}{ (x-2)(x+1)}= \\ \\ =\frac{ -1}{ 1}* \frac{ (x-3)}{ (x-2)}=- \frac{ x-3}{ x-2}$