Đáp án:

$\begin{array}{l}
a){x^2} - 1 - 4{y^2} + 4y\\
 = {x^2} - \left( {4{y^2} - 4y + 1} \right)\\
 = {x^2} - {\left( {2y - 1} \right)^2}\\
 = \left( {x - 2y + 1} \right)\left( {x + 2y - 1} \right)\\
b)2{x^2} - {y^2} + xy\\
 = 2{x^2} + 2xy - xy - {y^2}\\
 = \left( {x + y} \right)\left( {2x - y} \right)\\
c){x^2} - 20x + 84 = 0\\
 \Leftrightarrow {x^2} - 6x - 14x + 84 = 0\\
 \Leftrightarrow \left( {x - 6} \right)\left( {x - 14} \right) = 0\\
 \Leftrightarrow x = 6;x = 14\\
Vậy\,x = 6;x = 14\\
d)2xy + 4x + 2y + 1 > 5{x^2} + 2{y^2}\\
 \Leftrightarrow 5{x^2} + 2{y^2} - 2xy - 4x - 2y - 1 < 0\\
 \Leftrightarrow \left( {4{x^2} - 4x + 1} \right) + \left( {{x^2} - 2xy + {y^2}} \right)\\
 + {y^2} - 2y + 1 - 3 < 0\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} + {\left( {x - y} \right)^2} + {\left( {y - 1} \right)^2} < 3\\
Do:x;y \in Z\\
{\left( {2x - 1} \right)^2} + {\left( {x - y} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\\
 + TH1:\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 0\\
{\left( {x - y} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y\\
y = 1
\end{array} \right.\left( {ktm} \right)\\
 + TH2::\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 1\\
{\left( {x - y} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
x = y\\
y = 1
\end{array} \right. \Leftrightarrow x = y = 1\\
 + TH3:\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 0\\
{\left( {x - y} \right)^2} = 1\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y + 1/x = y - 1\\
y = 1
\end{array} \right.\left( {ktm} \right)\\
 + TH4::\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 0\\
{\left( {x - y} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y\\
y = 0/y = 2
\end{array} \right.\left( {ktm} \right)\\
 + TH5:\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 1\\
{\left( {x - y} \right)^2} = 1\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0/x = 1\\
x = y + 1/x = y - 1\\
y = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
 + TH6::\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 1\\
{\left( {x - y} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1/x = 0\\
x = y\\
y = 1/y = 2
\end{array} \right. \Leftrightarrow x = y = 1\\
 + TH7::\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 0\\
{\left( {x - y} \right)^2} = 1\\
{\left( {y - 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y + 1/x = y - 1\\
y = 1/y = 2
\end{array} \right.\left( {ktm} \right)\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;1} \right);\left( {0;1} \right)} \right\}\\
e){\left( {{x^2} + 4x} \right)^2} - 2\left( {{x^2} + 4x} \right) - 15 = 0\\
Dat:{x^2} + 4x = a\\
 \Leftrightarrow {a^2} - 2a - 15 = 0\\
 \Leftrightarrow {a^2} - 5a + 3a - 15 = 0\\
 \Leftrightarrow \left( {a - 5} \right)\left( {a + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x - 5 = 0\\
{x^2} + 4x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x + 5} \right) = 0\\
\left( {x + 1} \right)\left( {x + 3} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 5\\
x =  - 1\\
x =  - 3
\end{array} \right.\\
Vậy\,x \in \left\{ { - 5; - 3; - 1;1} \right\}
\end{array}$