Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

để được tổng =0 thì x + 2006/2007 = 0 và 2008/2009 - y =0

vậy suy ra x + 2006/2007 = 0 ; x = -2006/2007

suy ra 2008/2009 - y = 0 ; y = 2008/2009

Vì \(\left|x+\frac{2006}{2007}\right|\ge0;\left|\frac{2008}{2009}-y\right|\ge0\)

Mà \(\left|x+\frac{2006}{2007}\right|+\left|\frac{2008}{2009}-y\right|=0\)

=> \(\hept{\begin{cases}\left|x+\frac{2006}{2007}\right|=0\\\left|\frac{2008}{2009}-y\right|=0\end{cases}}\)=> \(\hept{\begin{cases}x+\frac{2006}{2007}=0\\\frac{2008}{2009}-y=0\end{cases}}\)=> \(\hept{\begin{cases}x=-\frac{2006}{2007}\\y=\frac{2008}{2009}\end{cases}}\)

Vì mũ chẵn và GTTĐ luôn lớn hơn hoặc bằng 0

mà ... ( ghi đề bài ra )

\(\Rightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\\\frac{4}{3}x+\frac{5}{2}y=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-4}{3}\end{cases}}\)

Vậy,.......

c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)

Nên x + 2009 = 0 => x = -2009

Ta có nhận xét : \(a+b=1\) thì

\(f\left(a\right)+f\left(b\right)=\frac{4^a}{4^a+2}+\frac{4^b}{4^b+2}=\frac{4^a\left(4^a+2\right)+4^b\left(4^b+2\right)}{\left(4^a+2\right)\left(4^b+2\right)}\)

                 \(=\frac{4^{a+b}+2.4^a+4^{a+b}+2.4^b}{4^{a+b}+2.4^a+2.4^b+4}=\frac{2.4^a+2.4^b+8}{2.4^a+2.4^b+8}=1\)

Áp dụng kết quả trên ta có :

\(S=\left[f\left(\frac{1}{2007}\right)+f\left(\frac{2006}{2007}\right)\right]+\left[f\left(\frac{2}{2007}\right)+f\left(\frac{2005}{2007}\right)\right]+...+\left[f\left(\frac{1003}{2007}\right)+f\left(\frac{1004}{2007}\right)\right]\)

Vâyu \(S=1+1+1+...+1=1003\) (có 1003 số hạng)

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

phần đầu mk thiếu điều kiện,bn tự bổ sung nha

a) \(A=x^{15}+3x^{14}+5\)

\(=x^{14}\left(x+3\right)+5\)

\(=x^{14}.0+5\)

= 5

b) x = -3 => x + 3 = 0

\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(=\left(x^{2006}.0+1\right)^{2007}\)

\(=1^{2007}=1\)

\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)

\(A=x^{14}.\left(x+3\right)+5\)

\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)

\(A=0+5\)

\(A=5\)        \(\text{Vậy }A=5\text{ với x+3=0}\)

\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)

\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)

\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=\left(x^{2006}.0+1\right)^{2007}\)

\(B=\left(0+1\right)^{2007}\)

\(B=1^{2007}\)

\(B=1\)           \(\text{Vậy }B=1\text{ với x=-3}\)

Ngu thì câm mồm đi