Đáp án:
$\begin{array}{l}
B1)\\
1){\left( {x + 3y} \right)^2} = {x^2} + 6xy + 9{y^2}\\
2){\left( {x - ab} \right)^2} = {x^2} - 2x.ab + {a^2}{b^2}\\
3)\left( {x - 6z} \right)\left( {x + 6z} \right)\\
= {x^2} - 36{z^2}\\
B2)\\
a)A = {\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {x + y + x - y} \right)\left( {x + y - x + y} \right)\\
= 2x.2y\\
= 4xy\\
b)B = {\left( {x + y} \right)^2} - 2\left( {x + y} \right)\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
= {\left( {x + y - x + y} \right)^2}\\
= {\left( {2y} \right)^2}\\
= 4{y^2}\\
c)C = {\left( {x + y} \right)^3} - {\left( {x - y} \right)^3} - 2{y^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
- {x^3} + 3{x^2}y - 3x{y^2} + {y^3} - 2{y^3}\\
= 6{x^2}y\\
B3)\\
A = {x^2} - x + 2\\
= {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4}\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4} > 0\\
\Leftrightarrow A > 0
\end{array}$
