Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.3xy( x-3y)( 1-y)\\ b.\ \left( 2-x+\frac{2}{5}\right)( x-1\\ c.\ ( 5( x+y) -3) .( 5( x+y) +3)\\ d.\ ( 3-4x) .\left( 9+12x+16x^{2}\right)\\ e.\ ( x+1)( x-2)( x+2)\\ f.\ 3ab\left( x^{2} -2x+a-25\right)\\ h.( x+y)( x+y-4ax)\\ i.\ \left( x-y^{3}\right)\left( x^{2} +xy^{3} +y^{3}\right)\left( x^{3} +y^{9}\right)\\ k.\ -( b-2)^{2}( a+b)^{2}\\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ 3x^{2} y-9xy^{2} +6x^{2} y^{2}\\ =3x^{2} y-3x^{2} y^{2} -9xy^{2} +9x^{2} y^{2}\\ =3x^{2} y( 1-y) -9xy^{2}( 1-y)\\ =\left( 3x^{2} y-9xy^{2}\right)( 1-y)\\ =3xy( x-3y)( 1-y)\\ b.\ 2( x-1) -x( x-1) -\frac{2}{5}( 1-x)\\ =\left( 2-x+\frac{2}{5}\right)( x-1)\\ c.\ 25( x+y)^{2} -9\\ =( 5( x+y) -3) .( 5( x+y) +3)\\ d.\ 27-64x^{3}\\ =3^{3} -( 4x)^{3}\\ =( 3-4x) .\left( 3^{2} +3.4x+( 4x)^{2}\right)\\ =( 3-4x) .\left( 9+12x+16x^{2}\right)\\ e.\ x^{3} +x^{2} -4x-4\\ =x^{3} -4x+x^{2} -4\\ =x\left( x^{2} -4\right) +x^{2} -4\\ =( x+1)( x-2)( x+2)\\ f.\ 3abx^{2} -6a^{2} bx+3a^{3} b-75ab\\ =\ 3ab\left( x^{2} -2x+a-25\right)\\ g.\ x^{2} -4y^{2} +4y-1\\ =( x-2y)( x+2y) +4y-1\\ Câu\ này\ không\ phân\ tích\ thành\ nhân\ tử\ được\\ h.\ x^{2} +2xy+y^{2} -4ax^{2} -4axy\\ =( x+y)^{2} -4ax( x+y)\\ =( x+y)( x+y-4ax)\\ i.\ x^{6} -y^{18}\\ =\left( x^{3}\right)^{2} -\left( y^{9}\right)^{2}\\ =\left( x^{3} -y^{9}\right)\left( x^{3} +y^{9}\right)\\ =\left( x-y^{3}\right)\left( x^{2} +xy^{3} +y^{3}\right)\left( x^{3} +y^{9}\right)\\ k.\ a^{2} +2ab-3b^{2} +4b-1\\ =a^{2} +2ab+b^{2} -4b^{2} +4b-1\\ =-( b-2)^{2}( a+b)^{2}\\ \end{array}$
