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?

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

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

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

=>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 \(x=2010\)

Tag thầy Lâm không :)???

đề ko có vấn đề nhỉ?

Giải:

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

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

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

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

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

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

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

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

\(\Leftrightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy ...

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

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

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

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

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

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

\(\Leftrightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

ta có (x+1/2009 +1) + ( x+3/2007 + 1)- (x+5/2005 +1) - (x+7/1993 + 1) = 0

=>(x +100/ 2009) + (x+100/2007) - (x+100/2005)-(x+100/1993)

=> (x +100) * (1/2009 + 1/2007+ 1/2005 + 1/1993) = 0

=> x = -100

Bạn cứ tinh ý để ý đến phần tử và mẫu cộng lại bằng 100. Khi bạn bỏ phần x + 100 ra thì còn lại như trên. Sau đó lược bỏ còn lại x = -100

Mạn phép mk không chép đề , mk làm luôn nhé

\(\dfrac{x+1}{2009}+1+\dfrac{x+3}{2007}+1=\dfrac{x+5}{2005}+1+\dfrac{x+7}{1993}+1\)

⇔ \(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2005}-\dfrac{x+2010}{1993}=0\)

⇔( x + 2010 )\(\left(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}\right)=0\)

Ta thấy : \(\dfrac{1}{2009}< \dfrac{1}{2007}< \dfrac{1}{2005}< \dfrac{1}{1993}\)

⇒ \(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}< 0\)

⇒ x + 2010 = 0

⇒ x = -2010

KL....

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

=>x-2010=0

hay x=2010