a)=1/2*2/3......*19/20

=1/20

b)=3/2*4/3......*2008/2007

=3/2007

a quên = 3/2008

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

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

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

Vì \(\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)\ne0\)

=> x-2010 =0

=> x =2010

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

\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )

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

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

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

Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)

2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

​\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)

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

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

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\right)=0\Leftrightarrow x=2010\)

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

=>x-2010=0

hay x=2010

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

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

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

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

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

=> \(x-2010=0\) 

                 \(x=2010\)

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

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

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

=>x-2010=0 (vì \(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)\(\dfrac{1}{2006}\)≠0)

=>x=2010

\(A=x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)

\(=x^6-2006x^5-x^5+2006x^4+x^4-2006x^3-x^3+2006x^2+x^2-2006x-x+2006+1\)

\(=x^5\left(x-2006\right)-x^4\left(x-2006\right)+x^3\left(x-2006\right)-x^2\left(x-2006\right)+x\left(x-2006\right)-\left(x-2006\right)+1\)

\(=\left(x^5-x^4+x^3-x^2+x-1\right)\left(x-2006\right)+1\)

Thay x = 2006

\(\Leftrightarrow A=1\)

Vậy A = 1 tại x = 2006

\(A=x^6-2007.x^5+2007.x^4-2007.x^3+2007.x^2-2007.x+2007\)

\(=x^6-\left(x+1\right).x^5+\left(x+1\right).x^4-...+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-...-x+1\)

\(=1\)

Sai đề rồi 

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

 Xét ta thấy \(2009\ne2008\ne2007\ne2006\)

Mà để cho \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

Thì \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}=0\)hay \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\)

Mà \(x-1\ne x-2\ne x-3\ne x-4\)Nên \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

Không thể bằng 0 được

Ta có \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\) Nên \(x-1=2009;x-2=2008;x-3=2007;x-4=2006\)

Suy ra \(x=2010\)P/S: Sở dĩ \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\)

được là bởi vì \(2009=2010-1\)và \(2008=2010-2\)và \(2007=2010-3\)và \(2006=2010-4\)