Answer:

\(5x+53=2xy+8y^2\)

\(\Rightarrow2\left(5x+53\right)=2\left(2xy+8y^2\right)\)

\(\Rightarrow10x+106=4xy+16y^2\)

\(\Rightarrow10x-4xy=16y^2-106\)

\(\Rightarrow x=\frac{16y^2-106}{10-4y}\)

\(\Rightarrow x=\frac{\left(16y^2-100\right)-6}{10-4y}\)

\(\Rightarrow x=\frac{-\left(10-4y\right)\left(4y+10\right)}{10-4y}-\frac{6}{10-4y}\)

\(\Rightarrow x=-4y-10-\frac{6}{10-4y}\)

Để cho x và y thuộc Z thì 6 chia hết cho 10 - 4y

\(\Rightarrow10-4y\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=1\\10-4y=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4y=9\left(l\right)\\4y=11\left(l\right)\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=2\\10-4y=-2\end{cases}}\Rightarrow\orbr{\begin{cases}4y=8\\4y=12\end{cases}}\Rightarrow\orbr{\begin{cases}y=2\Rightarrow x=-21\\y=3\Rightarrow x=-19\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=3\\10-4y=-3\end{cases}}\Rightarrow\orbr{\begin{cases}4y=7\left(l\right)\\4y=13\left(l\right)\end{cases}}\)

Trường hợp: \(\orbr{\begin{cases}10-4y=6\\10-4y=-6\end{cases}}\Rightarrow\orbr{\begin{cases}4y=4\\4y=16\end{cases}}\Rightarrow\orbr{\begin{cases}y=1\Rightarrow x=-15\\y=4\Rightarrow x=-25\end{cases}}\)

Ta có: \(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)

\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)

\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)

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

Mà \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}>0\) \(\Rightarrow x+2007=0\Rightarrow x=-2007\)