Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ x\left( 8x^{3} -3x+6\right)\\ b.\ -2x^{4} +3x^{2} +5\\ c.\ x-1\\ d.\ -x^{4} +4x^{2} -4\\ e.\ x+2\\ f.\ 5-x\\ g.\ 4x^{2} +6x+9\\ h.\ x-3\\ i.\ -a^{2} b^{2} +4abc-4c^{2} \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \left( 24x^{5} -9x^{3} +18x^{2}\right) :3x\\ =\frac{3x^{2}\left( 8x^{3} -3x+6\right)}{3x} =x\left( 8x^{3} -3x+6\right)\\ b.\ \left( 14x^{6} -21x^{4} -35x^{2}\right) :\left( -7x^{2}\right)\\ =\frac{-7x^{2}\left( -2x^{4} +3x^{2} +5\right)}{-7x^{2}} =-2x^{4} +3x^{2} +5\\ c.\ \left( x^{2} -2x+1\right) :( x-1)\\ =\frac{( x-1)^{2}}{x-1} =x-1\\ d.\ \left( x^{6} -6x^{4} +12x^{2} -8\right) :\left( 2-x^{2}\right)\\ =\frac{\left( x^{2}\right)^{3} -2^{3} +6x^{2}\left( 2-x^{2}\right)}{2-x^{2}}\\ =\frac{\left( x^{2} -2\right)\left( x^{4} +2x^{2} +4\right) +6x^{2}\left( 2-x^{2}\right)}{\left( 2-x^{2}\right)}\\ =\frac{\left( 2-x^{2}\right) .\left( -x^{4} -2x^{2} -4+6x^{2}\right)}{2-x^{2}} =-x^{4} +4x^{2} -4\\ e.\ \left( x^{2} +5x+6\right) :( x+3)\\ =\frac{x^{2} +3x+2x+6}{x+3} =\frac{x( x+3) +2( x+3)}{x+3}\\ =\frac{( x+2)( x+3)}{x+3} =x+2\\ f.\ \left( 25-x^{2}\right) :( x+5)\\ =\frac{( 5-x)( 5+x)}{x+5} =5-x\\ g.\ \left( 8x^{3} +27\right) :( 2x+3)\\ =\frac{( 2x)^{2} +3^{3}}{2x+3} =\frac{( 2x+3)\left( 4x^{2} +6x+9\right)}{2x+3} =4x^{2} +6x+9\\ h.\ \left( x^{2} -3x+xy-3y\right) :( x+y)\\ =\frac{x( x-3) +y( x-3)}{x+y} =\frac{( x+y)( x-3)}{x+y} =x-3\\ i.\ \left( a^{3} b^{3} -6a^{2} b^{2} c+12abc^{2} -8c^{3}\right) :( 2c-ab)\\ =\frac{( ab)^{3} -( 2c)^{3} +6abc( 2c-ab)}{2c-ab}\\ =\frac{( ab-2c)\left( a^{2} b^{2} +2abc+4c^{2}\right) +6abc( 2c-ab)}{2c-ab}\\ =\frac{( 2c-ab)\left( -a^{2} b^{2} -2abc-4c^{2} +6abc\right)}{2c-ab}\\ =-a^{2} b^{2} +4abc-4c^{2} \end{array}$