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x+x+1+x+2+.........................+x+2003=2004
(x+x+x+...................+x)+(1+2+3+...................+2003)=2004
2004x+2007006=2004
2004x=2004:2007006=2/2003
x=2/2003:2004
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1-\frac{x-3}{2002}+1-\frac{x-4}{2001}+1=0\)
\(\Rightarrow\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}< \frac{1}{2002};\frac{1}{2003}< \frac{1}{2001}\)\(\Rightarrow\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
\(\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
Vậy x = 2005
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
=>\(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1-\frac{x-3}{2002}-1=\frac{x-4}{2001}-1\)
=> \(\frac{x-1-2004}{2004}+\frac{x-2-2003}{2003}-\frac{x-3-2002}{2002}=\frac{x-4-2001}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}=\frac{x-2005}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Do \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
=> x - 2005 = 0 => x = 2005
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)