Đáp án:

$\begin{array}{l}
a)x\left( {x + 4} \right) - {x^2} - 6x = 10\\
 \Rightarrow {x^2} + 4x - {x^2} - 6x = 10\\
 \Rightarrow  - 2x = 10\\
 \Rightarrow x =  - 5\\
\text{Vậy}\,x =  - 5\\
b){\left( {{x^2} + 2x} \right)^2} - 2{x^2} - 4x = 3\\
 \Rightarrow {\left( {{x^2} + 2x} \right)^2} - 2\left( {{x^2} + 2x} \right) - 3 = 0\\
 \Rightarrow {\left( {{x^2} + 2x} \right)^2} - 3\left( {{x^2} + 2x} \right) + \left( {{x^2} + 2x} \right) - 3 = 0\\
 \Rightarrow \left( {{x^2} + 2x} \right)\left( {{x^2} + 2x - 3} \right) + \left( {{x^2} + 2x - 3} \right) = 0\\
 \Rightarrow \left( {{x^2} + 2x - 3} \right)\left( {{x^2} + 2x + 1} \right) = 0\\
 \Rightarrow \left( {x - 1} \right)\left( {x + 3} \right){\left( {x + 1} \right)^2} = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 3\\
x =  - 1
\end{array} \right.\\
\text{Vậy}\,x = 1;x =  - 1;x =  - 3\\
c){\left( {x + \dfrac{1}{2}} \right)^2} - \left( {x + \dfrac{1}{2}} \right)\left( {x + 6} \right) = 8\\
 \Rightarrow {x^2} + x + \dfrac{1}{4} - \left( {{x^2} + 6x + \dfrac{1}{2}x + 3} \right) - 8 = 0\\
 \Rightarrow {x^2} + x + \dfrac{1}{4} - {x^2} - 6x - \dfrac{1}{2}x - 3 - 8 = 0\\
 \Rightarrow  - \dfrac{{11}}{2}x - \dfrac{{43}}{4} = 0\\
 \Rightarrow x = \dfrac{{ - 43}}{4}.\dfrac{2}{{11}}\\
 \Rightarrow x = \dfrac{{ - 43}}{{22}}\\
\text{Vậy}\,x = \dfrac{{ - 43}}{{22}}\\
d)9{x^2} + 6x + 4{y^2} - 8y + 5 = 0\\
 \Rightarrow 9{x^2} + 6x + 1 + 4{y^2} - 8y + 4 = 0\\
 \Rightarrow {\left( {3x + 1} \right)^2} + 4\left( {{y^2} - 2y + 1} \right) = 0\\
 \Rightarrow {\left( {3x + 1} \right)^2} + 4{\left( {y - 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {3x + 1} \right)^2} \ge 0\\
4{\left( {y - 1} \right)^2} \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
3x + 1 = 0\\
y - 1 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x =  - \dfrac{1}{3}\\
y = 1
\end{array} \right.\\
\text{Vậy}\,x =  - \dfrac{1}{3};y = 1
\end{array}$