Đáp án:
$\begin{array}{l}
B1)\\
f){\left( {2x - 1} \right)^2} = 4{x^2} - 4x + 1\\
= 4{x^2} + \left( { - 4x} \right) + 1\\
g){\left( {2x - 3} \right)^2}\\
= 4{x^2} - 12x + 9\\
= 4{x^2} + \left( { - 12x} \right) + 9\\
h){\left( {\frac{1}{2} - x} \right)^2}\\
= \frac{1}{4} - x + {x^2}\\
= {x^2} + \left( { - x} \right) + \frac{1}{4}\\
e){\left( { - 3x + 2} \right)^2}\\
= 9{x^2} - 12x + 4\\
= 9{x^2} + \left( { - 12x} \right) + 4\\
B2)c){\left( {2x - \frac{1}{4}} \right)^2}\\
= 4{x^2} - 2x + \frac{1}{2}\\
= 4{x^2} + \left( { - 2x} \right) + \frac{1}{2}\\
f){\left( {x - y + z} \right)^2}\\
= {x^2} + {y^2} + {z^2} - 2xy + 2xz - 2yz\\
g){\left( {x - 2y + z} \right)^2}\\
= {x^2} + 4{y^2} + {z^2} - 4xy + 2xz - 4yz\\
h){\left( {2x - y + 3} \right)^2}\\
= 4{x^2} + {y^2} + 9 - 4xy + 12x - 6y\\
B3)c)16{x^2} - 8x + 1\\
= {\left( {4x} \right)^2} - 2.4x.1 + 1\\
= {\left( {4x - 1} \right)^2}\\
f){x^2} - 3x + \frac{9}{4}\\
= {x^2} - 2.x.\frac{3}{2} + \frac{9}{4}\\
= {\left( {x - \frac{3}{2}} \right)^2}\\
h)\frac{{{x^2}}}{4} - \frac{1}{2}x + \frac{1}{4}\\
= {\left( {\frac{x}{2} - \frac{1}{2}} \right)^2}\\
B4)a)A = {\left( {x - y} \right)^2} + {\left( {x + y} \right)^2}\\
= {x^2} - 2xy + {y^2} + {x^2} + 2xy + {y^2}\\
= 2\left( {{x^2} + {y^2}} \right)\\
b)B = {\left( {2a + b} \right)^2} - {\left( {2a - b} \right)^2}\\
= \left( {2a + b - 2a + b} \right)\left( {2a + b + 2a - b} \right)\\
= 2b.4a\\
= 8ab\\
c)C = {\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {x + y - x + y} \right)\left( {x + y + x - y} \right)\\
= 2x.2y\\
= 4xy\\
d)D = {\left( {2x - 1} \right)^2} - 2{\left( {2x - 3} \right)^2} + 4\\
= 4{x^2} - 4x + 1 - 2\left( {4{x^2} - 12x + 9} \right) + 4\\
= - 4{x^2} + 20x - 13
\end{array}$
