Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)

=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)

=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)

=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)

=> \(x^2-1=0\)

=> \(x^2=1\)

=> \(x=\pm1\)

Vậy phương trình có 2 nghiệm là x = 1, x = -1 .

Giải:

Ta có:

\(\dfrac{x+2002}{16}+\dfrac{x+2003}{15}+\dfrac{x+2004}{14}+\dfrac{x+2005}{13}+\dfrac{x+2006}{12}=-5\)

\(\Leftrightarrow\dfrac{x+2002}{16}+\dfrac{x+2003}{15}+\dfrac{x+2004}{14}+\dfrac{x+2005}{13}+\dfrac{x+2006}{12}+5=0\)

\(\Leftrightarrow\dfrac{x+2002}{16}+1+\dfrac{x+2003}{15}+1+\dfrac{x+2004}{14}+1+\dfrac{x+2005}{13}+1+\dfrac{x+2006}{12}+1=0\)

\(\Leftrightarrow\dfrac{x+2002+16}{16}+\dfrac{x+2003+15}{15}+\dfrac{x+2004+14}{14}+\dfrac{x+2005+13}{13}+\dfrac{x+2006+12}{12}=0\)

\(\Leftrightarrow\dfrac{x+2018}{16}+\dfrac{x+2018}{15}+\dfrac{x+2018}{14}+\dfrac{x+2018}{13}+\dfrac{x+2018}{12}=0\)

\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{16}+\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

Vì \(\dfrac{1}{16}+\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\ne0\)

\(\Leftrightarrow x+2018=0\)

\(\Leftrightarrow x=-2018\)

Vậy ...

Ta có :

\(x=2005\Rightarrow x+1=2006\)

Thay \(2006=x+1\) vào biểu thức trên ta được : 

\(x^{2005}-\left(x+1\right)x^{2004}+\left(x+1\right)x^{2003}-\left(x+1\right)x^{2002}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(=x^{2005}-x^{2005}+x^{2004}-x^{2004}+x^{2003}-...-x^3+x^2-x^2+x-1\)

\(=x-1\) mà \(x=2005\)

\(\Rightarrow x^{2005}-2006.x^{2004}+2006.x^{2003}-2006.x^{2002}+...-2006.x^2+2006x-1=2005-1=2004\)

2002/2001>:,2003/2002>1.....

CÓ 8 PHÂN SỐ MỖI PHÂN SỐ CÓ GIÁ TRỊ LỚN HƠN 1 VÂY TỔNG CỦA 8 PHÂN SỐ LỚN HƠN 1 SẼ LỚN HƠN 8.

Dể thôi

Tìm khoảng cách 2-1=1

số số hạng:(2006-1):1+1=2006

Tổng là:(2006+1)*2006:2=2013021

vậy S là 2013021

s là 2013021

Khoảng cách các số là:

       2-1=1

Có tất cả số số hạng là:

       (2006-1):1+1=2006(số hạng)

S =:

       (2006+1).2006:2=2013021

x=-2007

\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)

\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)

\(A>8\)