Giúp mk vs nhé!!!!!!!

Question

Giúp mk vs nhé!!!!!!!

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1 year 2020-11-27T08:32:31+00:00 1 Answers 47 views 0

$\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}$