Giúp mk vs nhé!!!!!!!

Đáp án:

$\begin{array}{l}
a)\dfrac{{x + 5}}{{2017}} – \dfrac{{x + 5}}{{2018}} + \dfrac{{x + 5}}{{2019}} – \dfrac{{x + 5}}{{2020}} = 0\\
 \Rightarrow \left( {x + 5} \right).\left( {\dfrac{1}{{2017}} – \dfrac{1}{{2018}} + \dfrac{1}{{2019}} – \dfrac{1}{{2020}}} \right) = 0\\
 \Rightarrow x + 5 = 0\\
 \Rightarrow x =  – 5\\
Vay\,x =  – 5\\
b)\dfrac{{x + 13}}{{2006}} + \dfrac{{x + 2006}}{{13}} + \dfrac{{x + 1}}{{2018}} + 3 = 0\\
 \Rightarrow \dfrac{{x + 13}}{{2006}} + 1 + \dfrac{{x + 2006}}{{13}} + 1 + \dfrac{{x + 1}}{{2018}} + 1 = 0\\
 \Rightarrow \dfrac{{x + 13 + 2006}}{{2006}} + \dfrac{{x + 2006 + 13}}{{13}} + \dfrac{{x + 1 + 2018}}{{2018}} = 0\\
 \Rightarrow \dfrac{{x + 2019}}{{2006}} + \dfrac{{x + 2019}}{{13}} + \dfrac{{x + 2019}}{{2018}} = 0\\
 \Rightarrow \left( {x + 2019} \right)\left( {\dfrac{1}{{2006}} + \dfrac{1}{{13}} + \dfrac{1}{{2018}}} \right) = 0\\
 \Rightarrow x + 2019 = 0\\
 \Rightarrow x =  – 2019\\
Vay\,x =  – 2019\\
c){x^2} – 3x > 0\\
 \Rightarrow x\left( {x – 3} \right) > 0\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x – 3 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x > 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x < 3
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x > 3\\
x < 0
\end{array} \right.\\
Vay\,x > 3\,hoac\,x < 0\\
d)Dkxd:\left\{ \begin{array}{l}
x \ne  – 3\\
x \ne  – 7\\
x \ne  – 10
\end{array} \right.\\
\dfrac{4}{{\left( {x + 3} \right)\left( {x + 7} \right)}} + \dfrac{3}{{\left( {x + 7} \right)\left( {x + 10} \right)}}\\
 = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow \dfrac{{\left( {x + 7} \right) – \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 7} \right)}} + \dfrac{{\left( {x + 10} \right) – \left( {x + 7} \right)}}{{\left( {x + 7} \right)\left( {x + 10} \right)}}\\
 = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow \dfrac{1}{{x + 3}} – \dfrac{1}{{x + 7}} + \dfrac{1}{{x + 7}} – \dfrac{1}{{x + 10}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow \dfrac{1}{{x + 3}} – \dfrac{1}{{x + 10}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow \dfrac{{x + 10 – \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 10} \right)}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow \dfrac{7}{{\left( {x + 3} \right)\left( {x + 10} \right)}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
 \Rightarrow x = 7\left( {tmdk} \right)\\
Vay\,x = 7
\end{array}$