\(\dfrac{x+10}{2003}+\dfrac{x+6}{2007}+\dfrac{x+12}{2001}+3=0\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BP
2
24 tháng 4 2018
\(\dfrac{x+10}{2003}+\dfrac{x+6}{2007}+\dfrac{x+12}{2001}+3=0\)
\(\Leftrightarrow\dfrac{x+10}{2003}+1+\dfrac{x+6}{2007}+1+\dfrac{x+12}{2001}+1=0\)
\(\Leftrightarrow\dfrac{x+10+2003}{2003}+\dfrac{x+6+2007}{2007}+\dfrac{x+12+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2013}{2003}+\dfrac{x+2013}{2007}+\dfrac{x+2013}{2001}=0\)
\(\Leftrightarrow\left(x+2003\right)\left(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
Vậy pt có nghiệm x = -2013
KT
24 tháng 4 2018
\(\dfrac{x+10}{2003}\)+\(\dfrac{x+6}{2007}\)+\(\dfrac{x+12}{2001}\)+3=0
<=> \(\dfrac{x+10}{2003}\)+1+\(\dfrac{x+6}{2007}\)+1+\(\dfrac{x+12}{2001}\)+1=0
<=> (\(\dfrac{x+10}{2003}\)+1) + (\(\dfrac{x+6}{2007}\)+1) + (\(\dfrac{x+12}{2001}\)+1)=0
<=> \(\dfrac{x+2013}{2003}\)+\(\dfrac{x+2013}{2007}\)+\(\dfrac{x+2013}{2001}\)=0
<=> (x+2013)(\(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\))=0
<=> x+2013=0( Vì \(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\)>0)
<=> x= -2013
Vậy S={-2013}
12 tháng 7 2023
=>\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)
=>\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2001}{4006}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)
=>\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)
=>1/(x+1)=1/2-2001/4006=1/2003
=>x+1=2003
=>x=2002
VT
2
18 tháng 4 2017
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}=\dfrac{x+34}{2003}+\dfrac{x+36}{2001}\)
\(\Leftrightarrow\dfrac{x+30}{2007}+1+\dfrac{x+32}{2005}+1=\dfrac{x+34}{2003}+1+\dfrac{x+36}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}=\dfrac{x+2037}{2003}+\dfrac{x+2037}{2001}\)
\(\Leftrightarrow\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}-\dfrac{x+2037}{2003}-\dfrac{x+2037}{2001}=0\)
\(\Leftrightarrow\left(x+2037\right)\left(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\right)=0\)
\(\Rightarrow x+2037=0\).Do \(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\ne0\)
\(\Rightarrow x=-2037\)
VT
19 tháng 4 2017
Các bạn xem mình làm thế này có đúng không nhé. Nếu sai thì xin các bạn chữa hộ mình
Bài làm
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}=\dfrac{x+34}{2003}+\dfrac{x+36}{2001}\)
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}-\dfrac{x+34}{2003}-\dfrac{x+36}{2001}=0\)
\(\left(\dfrac{x+30}{2007}+1\right)+\left(\dfrac{x+32}{2005}+1\right)-\left(\dfrac{x+34}{2003}+1\right)-\left(\dfrac{x+36}{2001}+1\right)=0\)
\(\dfrac{x+30+2007}{2007}+\dfrac{x+32+2005}{2005}-\dfrac{x+34+2003}{2003}-\dfrac{x+36+2001}{2001}=0\)\(\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}-\dfrac{x+2037}{2003}-\dfrac{x+2037}{2001}=0\)\(\left(x+2037\right).\left(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\right)=0\)
x+2037=0
x = -2037
18 tháng 9 2021
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) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
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) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
5 tháng 3 2021
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
AH
Akai Haruma
Giáo viên
5 tháng 3 2021
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
21 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
21 tháng 8 2017
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...
30 tháng 3 2017
\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005
30 tháng 3 2017
x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4
<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+
(x-2002/3 -1)+(x-2001/4 -1) =0
<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-
x-2005/3- x-2005/4 =0
<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0
<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)
<=>x=2005
Vậy pt có nghiệm là x=2005
\(\dfrac{x+10}{2003}+\dfrac{x+6}{2007}+\dfrac{x+12}{2001}+3=0\)
<=>\(\dfrac{x+10}{2003}+1+\dfrac{x+6}{2007}+1+\dfrac{x+12}{2001}+1=0\)
<=>\(\dfrac{x+2013}{2003}+\dfrac{x+2013}{2007}+\dfrac{x+2013}{2001}=0\)
<=>\(\left(x+13\right)\left(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\right)=0\)
vì 1/2003+1/2007+1/2001 khác 0
=>x+13=0<=>x=-13
vậy.............