Giải pt:
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
K
Khách
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SB
4
3 tháng 3 2017
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy x = -2010
3 tháng 3 2017
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\dfrac{x+2}{2008}+1+\dfrac{x+3}{2007}+1+\dfrac{x+4}{2006}+1+\dfrac{x+2028}{6}-3=0\)
\(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+2010=0\). Do \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x=-2010\)
TA
3
MS
6 tháng 3 2019
\(\frac{x+2}{2008}\)+ 1 + \(\frac{x+3}{2007}\)+1 +\(\frac{x+4}{2006}\)+1 +\(\frac{x+2028}{6}\)-3=0
\(\Leftrightarrow\)\(\frac{x+2+2008}{2008}\)+ \(\frac{x+3+2007}{2007}\) + \(\frac{x+4+2006}{2006}\)+ \(\frac{x+2028-18}{6}\)= 0
\(\Leftrightarrow\) \(\frac{x+2010}{2008}\)+ \(\frac{x+2010}{2007}\)+ \(\frac{x+2010}{2006}\)+ \(\frac{x+2010}{6}\)=0
\(\Leftrightarrow\)(x +2010).\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)\)=0
\(\Leftrightarrow\)x + 2010 = 0 \(\left(vì\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\right)\)
\(\Leftrightarrow\) x = -2010
Vậy S = \(\left\{-2010\right\}\)
BS
9 tháng 3 2018
⇔ \(\dfrac{x+2}{2008}\) +1 +\(\dfrac{x+3}{2007}\) +1+\(\dfrac{x+4}{2006}\)+1 +\(\dfrac{2028}{6}\)-3 =0
⇔\(\dfrac{x+2}{2008}+\dfrac{2008}{2008}+\dfrac{x+3}{2007}+\dfrac{2007}{2007}+\dfrac{x+4}{2006}+\dfrac{2006}{2006}+\dfrac{x+2028}{6}-\dfrac{18}{6}=0\)
⇔\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
⇔(x+2010)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
Mà \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)\)≠0
⇒x+2010=0
⇔x=-2010
Vậy phương trình có nghiệm x=-2010
DT
1
7 tháng 3 2017
Không chẳng có vấn đề gì cả. có thể sai so với cái đề nào đó "nội hàm nó đúng"
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{-x+4}{2006}+\dfrac{-x-2008}{6}\)
\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).x=\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)\)\(x=\dfrac{\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)}{\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).}\)
Thích thì rút gọn chẳng thích thì kệ nó
MV
12 tháng 8 2017
2.
\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
\(\Rightarrow x>0\)
Vậy \(x>0\)
HM
3
23 tháng 10 2023
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
=>x-2010=0
=>x=2010
KV
23 tháng 10 2023
(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006
(x - 1)/2009 - 1 + (x - 2)/2008 - 1 = (x - 3)/2007 - 1 + (x - 4)/2006 - 1
(x - 2010)/2009 + (x - 2010)/2008 = (x - 2010)/2007 + (x - 2010)/2006
(x - 2010)/2009 + (x - 2010)/2008 - (x - 2010)/2007 - (x - 2010)/2006 = 0
(x - 2010).(1/2009 + 1/2008 - 1/2007 - 1/2006) = 0
x - 2010 = 0
x = 2010
BT
2
19 tháng 4 2023
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)
mà `1/2008+1/2007+1/2005+1/2006≠ 0`
`=> x+2010=0`
`=>x=-2010`
19 tháng 4 2023
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)
=>x+2010=0
=>x=-2010
LE
1
25 tháng 2 2020
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\\ \Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\\ \Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\\ \Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\\ \Leftrightarrow x+2010=0\\ \Leftrightarrow x=-2010\)
Vậy pt có tập nghiệm \(S=\left\{-2010\right\}\)
AH
Akai Haruma
Giáo viên
28 tháng 1 2021
Lời giải:
a)
PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)
\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)
\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)
\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)
\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)
b) Bạn kiểm tra lại xem có sai đề không?
GH
1
17 tháng 7 2022
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)
\(\Leftrightarrow x+2010=0\) vì \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)
=> x=-2010
vậy.....