Đáp án:

$\begin{array}{l}
B1)\\
a){\left( {x + \frac{1}{6}y + 3} \right)^2}\\
 = {x^2} + {\left( {\frac{1}{6}y} \right)^2} + {3^2} + 2.x.\frac{1}{6}y\\
 + 2.\frac{1}{6}y.3 + 2.x.3\\
 = {x^2} + \frac{1}{{36}}{y^2} + 9 + \frac{1}{3}xy + y + 6x\\
b){\left( {2x + 3} \right)^2}.{\left( {x + 1} \right)^2}\\
 = {\left[ {\left( {2x + 3} \right).\left( {x + 1} \right)} \right]^2}\\
 = {\left( {2{x^2} + 5x + 3} \right)^2}\\
 = 4{x^4} + 25{x^2} + 9 + 2.2{x^2}.5x\\
 + 2.2{x^2}.3 + 2.5x.3\\
 = 4{x^4} + 25{x^2} + 9 + 20{x^3} + 12{x^2} + 30x\\
B2)\\
{\left( {3x + 1} \right)^2} - 9\left( {x + 2} \right) =  - 5\\
 \Rightarrow 9{x^2} + 6x + 1 - 9x - 18 =  - 5\\
 \Rightarrow 9{x^2} - 3x - 12 = 0\\
 \Rightarrow 3{x^2} - x - 4 = 0\\
 \Rightarrow \left( {3x - 4} \right)\left( {x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 1\\
x = \frac{4}{3}
\end{array} \right.\\
B3)\\
{\left( {a + b} \right)^2} = {a^2} + {b^2}\\
 \Rightarrow {a^2} + 2ab + {b^2} = {a^2} + {b^2}\\
 \Rightarrow 2ab = 0\\
 \Rightarrow \left[ \begin{array}{l}
a = 0\\
b = 0
\end{array} \right.
\end{array}$

Vậy a=0 hoặc b=0