Đáp án:
$\begin{array}{l}
a){\left( {\dfrac{1}{2} + x} \right)^2}\\
= {\left( {\dfrac{1}{2}} \right)^2} + 2.\dfrac{1}{2}.x + {x^2}\\
= {x^2} + x + \dfrac{1}{4}\\
{\left( {2x + 1} \right)^2} = 4{x^2} + 4x + 1\\
b){\left( {2x + 3y} \right)^2} = 4{x^2} + 12xy + 9{y^2}\\
{\left( {0,01 + xy} \right)^2}\\
= 0,{01^2} + 2.0,01.xy + {x^2}{y^2}\\
= {x^2}{y^2} + 0,02xy + 0,0001\\
c){\left( {\dfrac{1}{2} - x} \right)^2} = \dfrac{1}{4} - x + {x^2}\\
{\left( {2x - 1} \right)^2} = 4{x^2} - 4x + 1\\
d){\left( {2x - 3y} \right)^2} = 4{x^2} - 12xy + 9{y^2}\\
{\left( {0,01 - xy} \right)^2}\\
= 0,{01^2} - 2.0,01.xy + {x^2}{y^2}\\
= {x^2}{y^2} - 0,02xy + 0,0001\\
e)\left( {x + 1} \right)\left( {x - 1} \right)\\
= {x^2} - 1\\
f)\left( {x - 2y} \right)\left( {x - 2y} \right)\\
= {\left( {x - 2y} \right)^2}\\
= {x^2} - 2.x.2y + {\left( {2y} \right)^2}\\
= {x^2} - 4xy + 4{y^2}\\
56.64\\
= \left( {60 - 4} \right)\left( {60 + 4} \right)\\
= {60^2} - {4^2}\\
= 3600 - 16\\
g)\left( {x + y + z} \right)\left( {x - y - z} \right)\\
= \left( {x + y + z} \right)\left( {x - \left( {y + z} \right)} \right)\\
= {x^2} - {\left( {y + z} \right)^2}\\
= {x^2} - \left( {{y^2} + 2yz + {z^2}} \right)\\
= {x^2} - {y^2} - {z^2} - 2yz\\
h)\left( {x - y + z} \right)\left( {x + y + z} \right)\\
= \left( {x + z - y} \right)\left( {x + z + y} \right)\\
= {\left( {x + z} \right)^2} - {y^2}\\
= {x^2} + 2xz + {z^2} - {y^2}
\end{array}$
