$\frac{x}{x+1}=\frac{3}{x+1}$
$1\left(\frac{\left(x-2\right)\left(x-3\right)}{6}\right)+3\left(\frac{\left(x-0\right)\left(x-3\right)}{-2}\right)+\left(\frac{\left(x-0\right)\left(x-2\right)}{3}\right)$
$\:2q\:+\:1-2$
$0^2+2\cdot0\cdot3+3^2$
$-x^2+8x^2$
$-a+a+a-b-a-b+c+a+b$
$x^2-30x+169$
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