Đáp án:

$\begin{array}{l}
B1)\\
1){\left( {x + 5} \right)^2} - {\left( {x - 5} \right)^2}\\
 = \left( {x + 5 + x - 5} \right)\left( {x + 5 - x + 5} \right)\\
 = 2x.10\\
 = 20x\\
2){\left( {4x - 5} \right)^2} - \left( {4x + 5} \right)\left( {4x - 5} \right)\\
 = \left( {4x - 5} \right)\left( {4x - 5 - 4x - 5} \right)\\
 = \left( {4x - 5} \right).\left( { - 10} \right)\\
 = 50 - 40x\\
3){\left( {2x + 3y} \right)^2} - {\left( {2x - 3y} \right)^2}\\
 = \left( {2x + 3y + 2x - 3y} \right)\left( {2x + 3y - 2x + 3y} \right)\\
 = 4x.6y\\
 = 24xy\\
4)16\left( {x + 1} \right)\left( {x - 1} \right) - {\left( {5x + 1} \right)^2}\\
 = 16\left( {{x^2} - 1} \right) - 25{x^2} - 10x - 1\\
 =  - 9{x^2} - 10x - 17\\
5){\left( {x + 3} \right)^2} - \left( {x - 4} \right)\left( {x + 4} \right)\\
 = {x^2} + 6x + 9 - {x^2} + 16\\
 = 6x + 25\\
6)\left( {2x + 1} \right)\left( {2x - 1} \right) - {\left( {2x - 1} \right)^2}\\
 = \left( {2x - 1} \right)\left( {2x + 1 - 2x + 1} \right)\\
 = \left( {2x - 1} \right).2\\
 = 4x - 2\\
7){\left( {x + 1} \right)^2} - \left( {x - 2} \right)\left( {x + 2} \right) + 3x\\
 = {x^2} + 2x + 1 - {x^2} + 4 + 3x\\
 = 5x + 5\\
8){\left( {a + b - c} \right)^2} - \left( {a + b} \right)\left( {b - c} \right)\\
 = {a^2} + {b^2} + {c^2} + 2ab - 2ac - 2bc\\
 - \left( {ab - ac + {b^2} - bc} \right)\\
 = {a^2} + {c^2} + ab - ac - bc\\
41)\\
2){\left( {2x - 1} \right)^2} - 4{x^2} = 0\\
 \Leftrightarrow 4{x^2} - 4x + 1 - 4{x^2} = 0\\
 \Leftrightarrow  - 4x + 1 = 0\\
 \Leftrightarrow 4x = 1\\
 \Leftrightarrow x = \dfrac{1}{4}\\
Vậy\,x = \dfrac{1}{4}\\
3){\left( {x + 4} \right)^2} - x\left( {x + 2} \right) = 0\\
 \Leftrightarrow {x^2} + 8x + 16 - {x^2} - 2x = 0\\
 \Leftrightarrow 6x + 16 = 0\\
 \Leftrightarrow 6x =  - 16\\
 \Leftrightarrow x =  - \dfrac{8}{3}\\
Vậy\,x = \dfrac{{ - 8}}{3}\\
4){\left( {2 + 3x} \right)^2} - 9{x^2} = 0\\
 \Leftrightarrow 4 + 12x + 9{x^2} - 9{x^2} = 0\\
 \Leftrightarrow 12x + 4 = 0\\
 \Leftrightarrow x =  - \dfrac{1}{3}\\
Vậy\,x =  - \dfrac{1}{3}\\
5){x^2} - 25 = 0\\
 \Leftrightarrow {x^2} = 25\\
 \Leftrightarrow x = 5;x =  - 5\\
Vậy\,x = 5;x =  - 5\\
6)4{x^2} - 121 = 0\\
 \Leftrightarrow {x^2} = \dfrac{{121}}{4}\\
 \Leftrightarrow x = \dfrac{{11}}{2};x =  - \dfrac{{11}}{2}\\
Vậy\,x = \dfrac{{11}}{2};x =  - \dfrac{{11}}{2}\\
7)\dfrac{4}{9}{x^2} - 16 = 0\\
 \Leftrightarrow \dfrac{4}{9}{x^2} = 16\\
 \Leftrightarrow {x^2} = 36\\
 \Leftrightarrow x = 6;x =  - 6\\
Vậy\,x = 6;x =  - 6\\
8)\dfrac{9}{4} + 4{x^2} - 6x = 0\\
 \Leftrightarrow 16{x^2} - 24x + 9 = 0\\
 \Leftrightarrow {\left( {4x} \right)^2} - 2.4x.3 + {3^2} = 0\\
 \Leftrightarrow {\left( {4x - 3} \right)^2} = 0\\
 \Leftrightarrow 4x - 3 = 0\\
 \Leftrightarrow x = \dfrac{3}{4}\\
Vậy\,x = \dfrac{3}{4}
\end{array}$