Đáp án:
19.$x=2010$
20.$x=\dfrac73$
Giải thích các bước giải:
19.Ta có:
$\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}$
$\to \dfrac{x-2}{2008}-\dfrac{x-4}{2006}=\dfrac{x-3}{2007}-\dfrac{x-1}{2009}$
$\to \dfrac{2006(x-2)-2008(x-4)}{2008\cdot 2006}=\dfrac{2009(x-3)-2007(x-1)}{2007\cdot 2009}$
$\to \dfrac{-2x+4020}{2008\cdot 2006}=\dfrac{2x-4020}{2007\cdot 2009}$
$\to \dfrac{2x-4020}{2007\cdot 2009}-\dfrac{-2x+4020}{2008\cdot 2006}=0$
$\to \dfrac{2x-4020}{2007\cdot 2009}+\dfrac{2x-4020}{2008\cdot 2006}=0$
$\to (2x-4020)(\dfrac{1}{2007\cdot 2009}+\dfrac{1}{2008\cdot 2006})=0$
$\to 2x-4020=0$
$\to 2x=4020$
$\to x=2010$
20.ĐKXĐ: $x\ne 1,3,8, 20$
Ta có:
$\dfrac{2}{(x-1)(x-3)}+\dfrac{5}{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=\dfrac{-3}4$
$\to \dfrac{(x-1)-(x-3)}{(x-1)(x-3)}+\dfrac{(x-3)-(x-8)}{(x-3)(x-8)}+\dfrac{(x-8)-(x-20)}{(x-8)(x-20)}-\dfrac1{x-20}=\dfrac{-3}4$
$\to\dfrac1{x-3}-\dfrac1{x-1}+\dfrac1{x-8}-\dfrac1{x-3}+\dfrac1{x-20}-\dfrac1{x-8}-\dfrac1{x-20}=-\dfrac34$
$\to -\dfrac1{x-1}=-\dfrac34$
$\to \dfrac1{x-1}=\dfrac34$
$\to x-1=\dfrac43$
$\to x=\dfrac73$ thỏa mãn đkxđ
