Đáp án:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

$\begin{array}{l}
a)\dfrac{2}{{n - 1}} \in Z\\
 \Rightarrow \left( {n - 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
 \Rightarrow n \in \left\{ { - 1;0;2;3} \right\}\\
b)\dfrac{2}{{ - n + 2}} \in Z\\
 \Rightarrow \left( { - n + 2} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
 \Rightarrow  - n \in \left\{ { - 4; - 3; - 1;0} \right\}\\
 \Rightarrow n \in \left\{ {0;1;3;4} \right\}\\
c)\dfrac{{ - 3}}{{2n - 1}} \in Z\\
 \Rightarrow 2n - 1 \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow 2n \in \left\{ { - 2;0;2;4} \right\}\\
 \Rightarrow n \in \left\{ { - 1;0;1;2} \right\}\\
a)\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{{x + y}}{{2 + 5}} = \dfrac{{35}}{7} = 5\\
 \Rightarrow \left\{ \begin{array}{l}
x = 10\\
y = 25
\end{array} \right.\\
b)\dfrac{{x + 2}}{{y + 10}} = \dfrac{1}{5}\\
 \Rightarrow \dfrac{{x + 2}}{1} = \dfrac{{y + 10}}{5} = \dfrac{{3x + 6}}{3}\\
 = \dfrac{{y + 10 - 3x - 6}}{{5 - 3}} = \dfrac{{\left( {y - 3x} \right) + 4}}{2}\\
 = \dfrac{{2 + 4}}{2} = 3\\
 \Rightarrow \left\{ \begin{array}{l}
x + 2 = 3\\
y + 10 = 15
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 5
\end{array} \right.\\
Vậy\,x = 1;x = 5\\
c)\dfrac{x}{4} = \dfrac{y}{5} = \dfrac{{2x}}{8} = \dfrac{{2x - y}}{{8 - 5}} = \dfrac{{15}}{3} = 5\\
 \Rightarrow \left\{ \begin{array}{l}
x = 20\\
y = 25
\end{array} \right.
\end{array}$