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

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