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What is love?
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Huong San
31 tháng 7 2018 lúc 21:34

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)

\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)

\(\Leftrightarrow x+2010=0\)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)

=> x=-2010

vậy.....

Silver Bullet
Xem chi tiết
Nguyễn Huy Tú
3 tháng 3 2017 lúc 20:38

\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)

\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

\(\Rightarrow x+2010=0\)

\(\Rightarrow x=-2010\)

Vậy x = -2010

Lightning Farron
3 tháng 3 2017 lúc 20:39

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)

\(\Leftrightarrow\dfrac{x+2}{2008}+1+\dfrac{x+3}{2007}+1+\dfrac{x+4}{2006}+1+\dfrac{x+2028}{6}-3=0\)

\(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow x+2010=0\). Do \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\ne0\)

\(\Leftrightarrow x=-2010\)

Trần Thiên Kim
3 tháng 3 2017 lúc 20:40

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0+1+1+1-3\) \(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

Vậy S={-2010}

trần anh tú
Xem chi tiết
Min Shuu
6 tháng 3 2019 lúc 9:27

\(\frac{x+2}{2008}\)+ 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 = \(\left\{-2010\right\}\)

bullet sivel
9 tháng 3 2018 lúc 22:34

\(\dfrac{x+2}{2008}\) +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)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)

\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)\)≠0

⇒x+2010=0

⇔x=-2010

Vậy phương trình có nghiệm x=-2010

Thanh Trà
20 tháng 2 2018 lúc 20:29

Bạn tham khảo link này,định copy ra nhưng cơ mà thôi,tự xem nhé.

Câu hỏi của Silver Bullet - Toán lớp 8 | Học trực tuyến

hà my
Xem chi tiết
Nguyễn Lê Phước Thịnh
23 tháng 10 2023 lúc 12:34

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

=>x-2010=0

=>x=2010

Kiều Vũ Linh
23 tháng 10 2023 lúc 12:47

(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

Nguyễn Tuấn Tú
23 tháng 10 2023 lúc 13:04

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\\\Rightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-\dfrac{x-3}{2007}-\dfrac{x-4}{2006}=0\\\Rightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)-\left(\dfrac{x-3}{2007}-1\right)-\left(\dfrac{x-4}{2006}-1\right)=0 \\ \Rightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}-\dfrac{x-3-2007}{2007}-\dfrac{x-4-2006}{2006}=0\\ \Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\\\Rightarrow \left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\\ \)
Mà \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
\(\Rightarrow x-2010=0\\ \Rightarrow x=2010\)

Vậy \(x=2010\)

 

Bùi Thái Hà
Xem chi tiết
Thuỳ Linh Nguyễn
19 tháng 4 2023 lúc 15:00

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)

mà `1/2008+1/2007+1/2005+1/2006≠ 0`

`=> x+2010=0`

`=>x=-2010`

Nguyễn Lê Phước Thịnh
19 tháng 4 2023 lúc 14:32

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)

=>x+2010=0

=>x=-2010

Nguyễn Nhi
Xem chi tiết
Linh Phương
12 tháng 8 2017 lúc 8:51

Mở đầu về phương trình

Mở đầu về phương trình

Mới vô
12 tháng 8 2017 lúc 8:56

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)

George H. Dalton
Xem chi tiết
Nguyễn Lê Phước Thịnh
17 tháng 7 2022 lúc 13:44

\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

Thuy Khuat
Xem chi tiết
Giang
27 tháng 10 2017 lúc 21:36

Giải:

Ta có:

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

\(\Leftrightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-2=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}-2\)

\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)

\(\Leftrightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}=\dfrac{x-3-2007}{2007}+\dfrac{x-4-2006}{2006}\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)

Nên \(x-2010=0\)

\(\Rightarrow x=2010\)

Vậy \(x=2010\).

Chúc bạn học tốt!

NGUYEN MINH PHONG
27 tháng 10 2017 lúc 22:04
\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
chuyển vế ta có:
\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Rightarrow\left(x-10\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
\(\Rightarrow x-10=10\left(vi\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\right)\)\(\Rightarrow x=10\)
dfgfdgdfgdfgdfg
Xem chi tiết
Mỹ Duyên
30 tháng 3 2017 lúc 13:08

\(\dfrac{2-x}{2007}\) - 1 = \(\dfrac{1-x}{2008}\) - \(\dfrac{x}{2009}\)

<=> \(\dfrac{2-x}{2009}\) +1 -1 +1 = \(\dfrac{1-x}{2008}\) +1 - \(\dfrac{x}{2009}\) +1

<=> \(\dfrac{2-x+2007}{2007}\) = \(\dfrac{1-x+2008}{2008}\) + \(\dfrac{-x+2009}{2009}\)

<=> \(\dfrac{2009-x}{2007}\) = \(\dfrac{2009-x}{2008}\) + \(\dfrac{2009-x}{2009}\)

<=> (2009-x)(\(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) - \(\dfrac{1}{2009}\) ) = 0

<=> 2009 -x = 0

hoặc: \(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) -\(\dfrac{1}{2009}\) = 0

\(\dfrac{1}{2007}\) \(\ne\) \(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\)

=> \(\dfrac{1}{2007}\) - (\(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\) ) \(\ne\) 0

=> 2009 -x =0

<=> x =2009

Không Tên
30 tháng 3 2017 lúc 20:48

\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2009-x}{2007}-2=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}-2\)

\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)

\(\Rightarrow2009-x=0\Leftrightarrow x=2009\)