Sapo 976 Questions 1k Answers 0 Best Answers 0 Points View Profile0 Sapo Asked: Tháng Mười Một 27, 20202020-11-27T08:32:31+00:00 2020-11-27T08:32:31+00:00In: Môn ToánGiúp mk vs nhé!!!!!!!0Giúp mk vs nhé!!!!!!! ShareFacebookRelated Questions Где быстро занять денег? Một hình thang có đáy lớn là 52cm ; đáy bé kém đáy lớn 16cm ; chiều cao kém đáy ... Useful news and important articles1 AnswerOldestVotedRecentNeala 922 Questions 1k Answers 0 Best Answers 23 Points View Profile Neala 2020-11-27T08:34:16+00:00Added an answer on Tháng Mười Một 27, 2020 at 8:34 sáng 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}$0Reply Share ShareShare on FacebookLeave an answerLeave an answerHủy By answering, you agree to the Terms of Service and Privacy Policy .* Lưu tên của tôi, email, và trang web trong trình duyệt này cho lần bình luận kế tiếp của tôi.
Neala
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}$