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

⇔ \(\dfrac{x+2}{2008}\) +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

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

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

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

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

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

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

=> x=-2010

vậy.....

đề ko có vấn đề nhỉ?

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

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

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

=>x-2010=0

=>x=2010

(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

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

Vậy \(x=2010\)

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

mà `1/2008+1/2007+1/2005+1/2006≠ 0`

`=> x+2010=0`

`=>x=-2010`

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

=>x+2010=0

=>x=-2010

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

=>x-2010=0

hay x=2010

Giải:

Ta có:

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

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

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

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

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

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

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

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

Nên \(x-2010=0\)

\(\Rightarrow x=2010\)

Vậy \(x=2010\).

Chúc bạn học tốt!

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?

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

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

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

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

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

Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010

2010 sai chịu j cx chịu