Phân tích đa thức thành nhan tử a)x^2-2x-4y^2-4y b)x^4+2x^3-4x-4 c)x^2(1-x^2)-4-4x^2 d)(1+2x)(1-2x)-x(x+2)(x-2) e)x^2+y^2-x^2y^2+xy-x-y

\[\begin{array}{l}
a)\,\,\,{x^2} – 2x – 4{y^2} – 4y = \left( {{x^2} – 4{y^2}} \right) – \left( {2x + 4y} \right)\\
= \left( {x – 2y} \right)\left( {x + 2y} \right) – 2\left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {x + 2y – 2} \right).\\
b)\,\,{x^4} + 2{x^3} – 4x – 4 = \left( {{x^4} – 4} \right) + \left( {2{x^3} – 4x} \right)\\
= {\left( {{x^2}} \right)^2} – {2^2} + 2x\left( {{x^2} – 2} \right)\\
= \left( {{x^2} – 2} \right)\left( {{x^2} + 2} \right) + 2x\left( {{x^2} – 2} \right)\\
= \left( {{x^2} – 2} \right)\left( {{x^2} + 2x + 2} \right).\\
c)\,\,{x^2}\left( {1 – {x^2}} \right) – 4 + 4{x^2}\\
= {x^2}\left( {1 – {x^2}} \right) – 4\left( {1 – {x^2}} \right)\\
= \left( {1 – {x^2}} \right)\left( {{x^2} – 4} \right)\\
= \left( {1 – x} \right)\left( {1 + x} \right)\left( {x – 2} \right)\left( {x – 2} \right).\\
d)\,\,\left( {1 + 2x} \right)\left( {1 – 2x} \right) – x\left( {x + 2} \right)\left( {x – 2} \right)\\
= 1 – {\left( {2x} \right)^2} – x\left( {{x^2} – 4} \right)\\
= 1 – 4{x^2} – {x^3} + 4x\\
= \left( {1 – {x^3}} \right) – 4x\left( {x – 1} \right)\\
= \left( {1 – x} \right)\left( {{x^2} + x + 1} \right) + 4x\left( {1 – x} \right)\\
= \left( {1 – x} \right)\left( {{x^2} + x + 1 + 4x} \right)\\
= \left( {1 – x} \right)\left( {{x^2} + 5x + 1} \right).\\
e)\,\,\,{x^2} + {y^2} – {x^2}{y^2} + xy – x – y\\
= {x^2} – {x^2}{y^2} + \left( {xy – x} \right) + \left( {{y^2} – y} \right)\\
= {x^2}\left( {1 – {y^2}} \right) + x\left( {y – 1} \right) + y\left( {y – 1} \right)\\
= {x^2}\left( {1 – y} \right)\left( {1 + y} \right) + y\left( {y – 1} \right)\left( {x + y} \right)\\
= \left( {y – 1} \right)\left[ {y\left( {x + y} \right) – {x^2}\left( {1 + y} \right)} \right]\\
= \left( {y – 1} \right)\left( {xy + {y^2} – {x^2} – {x^2}y} \right).
\end{array}\]