+14915 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)]
\(\large \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\frac{2x+3}{2x-3}\) \(x\in \mathbb R\ |\ \ x\notin \{-1.5\ ;-1\}\)
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