Đáp án:
$\begin{array}{l}
B3)\\
a){x^2} - 4{x^2}{y^2} + {y^2} + 2xy\\
= {x^2} + 2xy + {y^2} - 4{x^2}{y^2}\\
= {\left( {x + y} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {x + y + 2xy} \right)\left( {x + y - 2xy} \right)\\
b)49 - {a^2} + 2ab - {b^2}\\
= {7^2} - \left( {{a^2} - 2ab + {b^2}} \right)\\
= {7^2} - {\left( {a - b} \right)^2}\\
= \left( {7 - a + b} \right)\left( {7 + a - b} \right)\\
c){a^2} - {b^2} + 4bc - 4{c^2}\\
= {a^2} - \left( {{b^2} - 4bc + 4{c^2}} \right)\\
= {a^2} - {\left( {b - 2c} \right)^2}\\
= \left( {a - b + 2c} \right)\left( {a + b - 2c} \right)\\
d)4{b^2}{c^2} - {\left( {{b^2} + {c^2} - {a^2}} \right)^2}\\
= \left( {2bc - {b^2} - {c^2} + {a^2}} \right)\left( {2bc + {b^2} + {c^2} - {a^2}} \right)\\
= \left( {{a^2} - {{\left( {b - c} \right)}^2}} \right)\left[ {{{\left( {b + c} \right)}^2} - {a^2}} \right]\\
= \left( {a - b + c} \right)\left( {a + b - c} \right)\left( {b + c - a} \right)\left( {b + c + a} \right)\\
B4)\\
a)\left( {{x^3} - 4{x^2}} \right) - \left( {x - 4} \right) = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
{x^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 1\\
x = - 1
\end{array} \right.\\
Vậy\,x = 4;x = 1;x = - 1\\
b){x^5} - 9x = 0\\
\Leftrightarrow x\left( {{x^4} - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^4} = 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\\
Vậy\,x = 0;x = \sqrt 3 ;x = - \sqrt 3
\end{array}$
