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Plzzz write all the steps

 

 Aug 16, 2020
 #1
avatar+10834 
+1

Subtract: \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)       [ \(2x^2-x-3=0\)   Assume: x = -1]

 

\((2x^2-x-3):(x+1)=2x-3\)

 \(\underline{2x^2+2x}\)

        \(-3x-3\)

        \(\underline{-3x-3}\)

                     \(0\)

 \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\) \(\frac{2x-3}{2x-3}\cdot \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)\(=\frac{2x^2-4x-3x+6-(3-12x)}{2x^2-x-3}\)     [expand with  (2x-3)]

\(=\frac{2x^2+5x+3}{2x^2-x-3}\)

 

  \((2x^2+5x+3):(x+1)=2x+3\)

   \(\underline{2x^2+2x}\)

               \(3x+3\)

               \(\underline{3x+3}\)

                         \(0\)

 

  \(\frac{2x^2+5x+3}{2x^2-x-3}\cdot\frac{x+1}{x+1}=\frac{2x+3}{2x-3}\)    [ shorten with (x+1)]

 

  \(\large \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\frac{2x+3}{2x-3}\)     \(x\in \mathbb R\ |\ \ x\notin \{-1.5\ ;-1\}\)

 

laugh  !

 Aug 16, 2020
edited by asinus  Aug 16, 2020
edited by asinus  Aug 16, 2020
edited by asinus  Aug 16, 2020

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