Giải phương trình, mọi người giúp mình với

Đáp án:

10) `x=-2009`

11) `x=2009`

Giải thích các bước giải:

$\begin{array}{l} 10)\dfrac{{x + 9}}{{2000}} + \dfrac{{x + 11}}{{1998}} = \dfrac{{x + 13}}{{1996}} + \dfrac{{x + 15}}{{1994}}\\ \Leftrightarrow \dfrac{{x + 9}}{{2000}} + 1 + \dfrac{{x + 11}}{{1998}} + 1 - \left( {\dfrac{{x + 13}}{{1996}} + 1} \right) - \left( {\dfrac{{x + 15}}{{1994}} + 1} \right) = 0\\ \Leftrightarrow \dfrac{{x + 2009}}{{2000}} + \dfrac{{x + 2009}}{{1998}} - \dfrac{{x + 2009}}{{1996}} - \dfrac{{x + 2009}}{{1994}} = 0\\ \Leftrightarrow (x + 2009)\left( {\dfrac{1}{{2000}} + \dfrac{1}{{1998}} - \dfrac{1}{{1996}} - \dfrac{1}{{1994}}} \right) = 0\\ \Leftrightarrow x + 2009 = 0\\ \Leftrightarrow x = - 2009\\ 11)\dfrac{{x - 10}}{{1999}} + \dfrac{{x - 20}}{{1989}} = \dfrac{{x - 50}}{{1959}} + \dfrac{{x - 80}}{{1929}}\\ \Leftrightarrow \dfrac{{x - 10}}{{1999}} - 1 + \dfrac{{x - 20}}{{1989}} - 1 - \left( {\dfrac{{x - 50}}{{1959}} - 1} \right) - \left( {\dfrac{{x - 80}}{{1929}} - 1} \right) = 0\\ \Leftrightarrow \dfrac{{x - 2009}}{{1999}} + \dfrac{{x - 2009}}{{1989}} - \dfrac{{x - 2009}}{{1959}} - \dfrac{{x - 2009}}{{1929}} = 0\\ \Leftrightarrow (x - 2009)\left( {\dfrac{1}{{1999}} + \dfrac{1}{{1989}} - \dfrac{1}{{1959}} - \dfrac{1}{{1929}}} \right) = 0\\ \Leftrightarrow x - 2009 = 0\\ \Leftrightarrow x = 2009 \end{array}$