..........We have:

But \(x;y\in Z+\) => \(\left(x;y\right)\in\left\{\left(3;8\right);\left(4;5\right);\left(5;4\right);\left(8;3\right)\right\}\)

Oh I'm sorry. I got the answer wrong. I'm going to answer this question again :(

Hey Dung! I'm My

\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\dfrac{3}{2}\Leftrightarrow1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x+y+1}{xy}1=\dfrac{1}{2}\Leftrightarrow2x+2y+2=xy\Leftrightarrow2x+2y+2-xy=0\)

\(\Leftrightarrow\left(2x-xy\right)-\left(4-2y\right)+6=0\Leftrightarrow x\left(2-y\right)-2\left(2-y\right)=-6\)

\(\Leftrightarrow\left(x-2\right)\left(2-y\right)=-6\)

We have:

But \(x;y\in Z+\) => (x;y)\(\in\left\{\left(1;8\right);\left(8;1\right)\right\}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{1}{100.103}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-2016}{2017}=\dfrac{1}{2017}\)