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a)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
b)\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(1+\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}=1+\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}\)
\(\Rightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
Giải tương tự câu a ta được \(x=-2018\)
a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow6006\left(x+1\right)+5460\left(x+1\right)+5005\left(x+1\right)=4620\left(x+1\right)+4290\left(x+1\right)\)
\(\Leftrightarrow\left(6006+5460+5005\right)\cdot\left(x+1\right)=\left(4620+4290\right)\cdot\left(x+1\right)\)
\(\Leftrightarrow16471\left(x+1\right)=8910\left(x+1\right)\)
\(\Leftrightarrow16471x+16471=8910x+8910\)
\(\Leftrightarrow16471x-8910x=8910-16471\)
\(\Leftrightarrow7561x=-7561\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
b) \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Rightarrow4096749040\left(x+4\right)+4094735904\left(x+3\right)=4092704785\left(x+2\right)+4090675680\left(x+1\right)\)
\(\Leftrightarrow4096769040x+16387076160+4094735904x+12284207712=4092704785x+8185409570+4090675680x+4090675680\)
\(\Leftrightarrow8191504944x+28671283872=8183380465x+12276085250\)
\(\Leftrightarrow8191504944x-8183380465x=12276085250-28671283872\)
\(\Leftrightarrow8124479x=-16395198622\)
\(\Rightarrow x=-2018\)
Vậy \(x=-2017\)
P/s: đây không phải cách làm tối ưu, vì vậy mình nghĩ bạn nên tham khảo từ các bài làm khác nhé!
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\dfrac{x-1}{2017}+\dfrac{x-2}{2016}-\dfrac{x-3}{2015}-\dfrac{x-4}{2014}=0\)
\(\Rightarrow\dfrac{x-1}{2017}-1+\dfrac{x-2}{2016}-1-\dfrac{x-3}{2015}+1-\dfrac{x-4}{2014}+1=0\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)-\left(\dfrac{x-3}{2015}-1\right)-\left(\dfrac{x-4}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Rightarrow x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Để \(x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x-2018=0\)
\(x=2018\)
Ta có :
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\) ( trừ 2 vế cho 2 )
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
=> x+1=0
<=> x=-1
b) \(\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
<=> \(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
đến đây tương tự a
a) Ta có:
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(Vì:\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
Vậy....
b)Sửa lại đề nha
Ta có:
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
Lý giải tương tự câu a và kết luận nha
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
\(=>\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)
=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\) = 0
Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)
=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)
=> \(x+2018=0\)
=>x = 0-2018
=> x = 0+(-2018)
=> x = -2018
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy...
\(\dfrac{x+4}{2001}+\dfrac{x+3}{2002}=\dfrac{x+2}{2003}+\dfrac{x+1}{2004}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2001}+1\right)+\left(\dfrac{x+3}{2002}+1\right)=\left(\dfrac{x+2}{2003}+1\right)+\left(\dfrac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\dfrac{x+2005}{2001}+\dfrac{x+2005}{2002}-\dfrac{x+2005}{2003}-\dfrac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right)\cdot\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)
Mà \(\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2005=0\Rightarrow x=-2005\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}-\dfrac{x+2018}{2016}-\dfrac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow xx+2018=0\Leftrightarrow x=-2018\)
Vậy x = -2018
Nguyễn Huy Tú, cho mk hỏi sao câu a bt đó lại bằng 0 vậy ? Mk ko hiểu lắm