Đáp án:

$\begin{array}{l}
1)Dkxd:x\# 5\\
\dfrac{{x - 3}}{{5 - x}} = \dfrac{5}{7}\\
 \Leftrightarrow 7.\left( {x - 3} \right) = 5.\left( {5 - x} \right)\\
 \Leftrightarrow 7x - 21 = 25 - 5x\\
 \Leftrightarrow 12x = 25 + 21\\
 \Leftrightarrow 12x = 46\\
 \Leftrightarrow x = \dfrac{{23}}{6}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{23}}{6}\\
b)x\# 1;x\#  - 7\\
\dfrac{{x - 2}}{{x - 1}} = \dfrac{{x + 4}}{{x + 7}}\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 7} \right) = \left( {x + 4} \right)\left( {x - 1} \right)\\
 \Leftrightarrow {x^2} + 5x - 14 = {x^2} + 3x - 4\\
 \Leftrightarrow 5x - 3x = 14 - 4\\
 \Leftrightarrow 2x = 10\\
 \Leftrightarrow x = 5\left( {tmdk} \right)\\
Vậy\,x = 5\\
c)\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{{2x}}{4} = \dfrac{{2x - y}}{{4 - 5}} = \dfrac{3}{{ - 1}} =  - 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 6\\
y =  - 15
\end{array} \right.\\
Vậy\,x =  - 6;y =  - 15\\
d)\dfrac{x}{2} = \dfrac{y}{5} = k \Leftrightarrow \left\{ \begin{array}{l}
x = 2k\\
y = 5k
\end{array} \right.\\
x.y = 10\\
 \Leftrightarrow 2k.5k = 10\\
 \Leftrightarrow {k^2} = 1\\
 \Leftrightarrow k = 1;k =  - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2;y = 5\\
x =  - 2;y =  - 5
\end{array} \right.\\
B2)\\
x:y:z = 3:4:5\\
 \Leftrightarrow \dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{5} = k \Leftrightarrow \left\{ \begin{array}{l}
x = 3k\\
y = 4k\\
z = 5k
\end{array} \right.\\
Do:2{x^2} + 2{y^2} - 3{z^2} =  - 100\\
 \Leftrightarrow 2.{\left( {3k} \right)^2} + 2.{\left( {4k} \right)^2} - 3.{\left( {5k} \right)^2} =  - 100\\
 \Leftrightarrow  - 25{k^2} =  - 100\\
 \Leftrightarrow {k^2} = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
k = 2 \Leftrightarrow \left\{ \begin{array}{l}
x = 6\\
y = 8\\
z = 10
\end{array} \right.\\
k =  - 2 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 6\\
y =  - 8\\
z =  - 10
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y;z} \right) = \left\{ {\left( {6;8;10} \right);\left( { - 6; - 8; - 10} \right)} \right\}
\end{array}$