Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x + 2} \right)\left( {3{x^2} - 4x} \right)\\
= 3{x^3} - 4{x^2} + 6{x^2} - 8x\\
= 3{x^3} + 2{x^2} - 8x\\
b,\\
\left( {{x^3} + 3{x^2} - 8x - 20} \right):\left( {x + 2} \right)\\
= \left[ {\left( {{x^3} + 2{x^2}} \right) + \left( {{x^2} + 2x} \right) - \left( {10x + 20} \right)} \right]:\left( {x + 2} \right)\\
= \left[ {{x^2}\left( {x + 2} \right) + x.\left( {x + 2} \right) - 10\left( {x + 2} \right)} \right]:\left( {x + 2} \right)\\
= \left[ {\left( {x + 2} \right)\left( {{x^2} + x - 10} \right)} \right]:\left( {x + 2} \right)\\
= {x^2} + x - 10\\
c,\\
\left( {2x - 3} \right)\left( {2x + 3} \right) - {\left( {x + 5} \right)^2} - \left( {x - 1} \right)\left( {x + 2} \right)\\
= \left( {4{x^2} - 9} \right) - \left( {{x^2} + 10x + 25} \right) - \left( {{x^2} + x - 2} \right)\\
= 4{x^2} - 9 - {x^2} - 10x - 25 - {x^2} - x + 2\\
= 2{x^2} - 11x - 32\\
d,\\
\left( {24{x^2}{y^3}{z^2} - 12{x^3}{y^2}{z^3} + 36{x^2}{y^2}{z^2}} \right):\left( { - 6{x^2}{y^2}{z^2}} \right)\\
= \left[ {6{x^2}{y^2}{z^2}.\left( {4y - 2x + 6} \right)} \right]:\left( { - 6{x^2}{y^2}{z^2}} \right)\\
= - \left( {4y - 2x + 6} \right)\\
= 2x - 4y + 6\\
a,\\
25 - {x^2} + 4xy - 4{y^2}\\
= {5^2} - \left( {{x^2} - 4xy + 4{y^2}} \right)\\
= {5^2} - {\left( {x - 2y} \right)^2}\\
= \left( {5 - x + 2y} \right)\left( {5 + x - 2y} \right)\\
b,\\
xy + xz - 2y - 2z\\
= \left( {xy + xz} \right) - \left( {2y + 2z} \right)\\
= x.\left( {y + z} \right) - 2.\left( {y + z} \right)\\
= \left( {y + z} \right)\left( {x - 2} \right)
\end{array}\)