wie sieht hier der gesamte lösungsweg aus?:

wie sieht hier der gesamte lösungsweg aus?:

(((2*x+1))/((x-1)))+(((2*x+4))/((1-x)))+(((x-9))/((x**2-1)))=(((3-8*x))/((1-x**2)))

$\begin{array}{rcl} \dfrac{ 2 x+1 }{ x-1 } +\dfrac{ 2x+4 }{ 1-x } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 }{ x-1 } -\dfrac{ 2x+4 }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 -(2x+4) }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ 2x+1 - 2x - 4) }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3 }{ x-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 } \qquad | \qquad x^2-1 = (x-1)(x+1)\\\\ \dfrac{ ( -3 ) }{( x-1 )}\cdot \dfrac{(x+1)}{(x+1)} +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ ( -3 )(x+1) }{ x^2-1 } +\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3x-3 }{ x^2-1 }+\dfrac{ x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -3x-3 +x-9 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -2x-12 }{ x^2-1 } &=&\dfrac{ 3-8x }{ 1-x^2 }\\\\ \dfrac{ -2x-12 }{ x^2-1 } &=&-\dfrac{ 3-8x }{ x^2-1 }\\\\ -\dfrac{ 2x+12 }{ x^2-1 } &=&-\dfrac{ 3-8x }{ x^2-1 }\\\\ \dfrac{ 2x+12 }{ x^2-1 } &=&\dfrac{ 3-8x }{ x^2-1 }\\\\ \dfrac{ 2x+12 }{ x^2-1 }-\dfrac{ 3-8x }{ x^2-1 } &=&0\\\\ \dfrac{ 2x+12-( 3-8x ) }{ x^2-1 } &=&0\\\\ \dfrac{ 2x+12-3+8x }{ x^2-1 } &=&0\\\\ \dfrac{ 10x+9 }{ x^2-1 } &=&0\\\\ 10x+9 &=&0\\ 10x &=&-9\\ x &=& \dfrac{-9}{10}\\ \mathbf{x} &\mathbf{=}& \mathbf{-0,9} \end{array}$