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Đáp án:

19) $S=\{2010\}$

20) $S=\left\{\dfrac{7}{3}\right\}$

Giải thích các bước giải:

19)

$\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\\⇔\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\\⇔\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\\⇔\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\\⇔(x-2010)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0$

Vì $\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0$

$\to x-2010=0\\⇔x=2010$

Vậy $S=\{2010\}$

20)

$\dfrac{2}{(x-1)(x-3)}+\dfrac{5}{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\,\,\,(x\ne1;x\ne3;x\ne8;x\ne20)\\⇔\dfrac{1}{x-3}-\dfrac{1}{x-1}+\dfrac{1}{x-8}-\dfrac{1}{x-3}+\dfrac{1}{x-20}-\dfrac{1}{x-8}-\dfrac{1}{x-20}=-\dfrac{3}{4}\\⇔-\dfrac{1}{x-1}=-\dfrac{3}{4}\\\to 3(x-1)=4\\⇔3x=7\\⇔x=\dfrac{7}{3}$

Vậy $S=\left\{\dfrac{7}{3}\right\}$