Đáp án:

$\begin{array}{l}
1)a)49 + \left( {11 - 25} \right)\\
 = 49 + 11 - 25\\
 = 60 - 25\\
 = 35\\
b) - 8 + 5.\left( { - 9} \right)\\
 =  - 8 - 45\\
 =  - 53\\
e)40 - {\left( { - 7} \right)^2}\\
 = 40 - 49\\
 =  - 9\\
f)\left| { - 15 + 21} \right| - \left| {4 - 11} \right|\\
 = \left| 6 \right| - \left| { - 7} \right|\\
 = 6 - 7\\
 =  - 1\\
b)5 + \left( { - 8} \right).3\\
 = 5 - 24\\
 =  - 19\\
d)4 + {\left( { - 5} \right)^2}\\
 = 4 + 25\\
 = 29\\
g)1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ...\\
 + 801 - 802 - 803 + 804\\
 = \left( {1 - 2 - 3 + 4} \right) + \left( {5 - 6 - 7 + 8} \right) + ... + \\
\left( {801 - 802 - 803 + 804} \right)\\
 = 0 + 0 + ... + 0\\
 = 0\\
B2)\\
 - 3 < x < 2\\
 \Rightarrow x \in \left\{ { - 2; - 1;0;1} \right\}\\
 \Rightarrow S = \left( { - 2} \right) + \left( { - 1} \right) + 0 + 1\\
 =  - 2\\
B3)a)x + 9 = 2 - 17\\
 \Rightarrow x = 2 - 17 - 9\\
 \Rightarrow x =  - 24\\
Vậy\,x =  - 24\\
c)x - 17 = \left( { - 11} \right).\left( { - 5} \right)\\
 \Rightarrow x = 17 + 55\\
 \Rightarrow x = 72\\
Vậy\,x = 72\\
d)\left| {x - 5} \right| = {\left( { - 4} \right)^2}\\
 \Rightarrow \left| {x - 5} \right| = 16\\
 \Rightarrow \left[ \begin{array}{l}
x - 5 = 16 \Rightarrow x = 21\\
x - 5 =  - 16 \Rightarrow x =  - 11
\end{array} \right.\\
Vậy\,x = 21;x =  - 11\\
b)x - 2 =  - 6 + 17\\
 \Rightarrow x = 2 - 6 + 17\\
 \Rightarrow x = 13\\
Vậy\,x = 13\\
e)2x + 5 = x - 1\\
 \Rightarrow 2x - x =  - 1 - 5\\
 \Rightarrow x =  - 6\\
Vậy\,x =  - 6\\
f)\left| {x - 4} \right| = \left| { - 81} \right|\\
 \Rightarrow \left| {x - 4} \right| = 81\\
 \Rightarrow \left[ \begin{array}{l}
x - 4 = 81 \Rightarrow x = 85\\
x - 4 =  - 81 \Rightarrow x = 77
\end{array} \right.\\
Vậy\,x = 85;x = 77\\
B4)a) - 7 \vdots x + 8\\
 \Rightarrow \left( {x + 8} \right) \in \left\{ { - 7; - 1;1;7} \right\}\\
 \Rightarrow x \in \left\{ { - 15; - 8; - 7; - 1} \right\}\\
b)\left( {3x - 13} \right) \vdots \left( {x - 2} \right)\\
3x - 13 = 3x - 6 - 7 = 3\left( {x - 2} \right) - 7\\
 \Rightarrow 7 \vdots \left( {x - 2} \right)\\
 \Rightarrow \left( {x - 2} \right) \in \left\{ { - 7; - 1;1;7} \right\}\\
 \Rightarrow x \in \left\{ { - 5;1;3;9} \right\}
\end{array}$

$\begin{array}{l}
B5)\left( {x - 2} \right)\left( {y + 1} \right) = 23 = 1.23\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 = 1\\
y + 1 = 23
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 =  - 1\\
y + 1 =  - 23
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 = 23\\
y + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 =  - 23\\
y + 1 =  - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 3;y = 22\\
x = 1;y =  - 24\\
x = 25;y = 0\\
x =  - 21;y =  - 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {3;22} \right);\left( {1; - 2} \right);\left( {25;0} \right);\left( { - 21; - 2} \right)} \right\}
\end{array}$