Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = a{b^3}{c^2} - {a^2}{b^2}{c^2} + a{b^2}{c^3} - {a^2}b{c^3}\\
= \left( {a{b^3}{c^2} - {a^2}{b^2}{c^2}} \right) + \left( {a{b^2}{c^3} - {a^2}b{c^3}} \right)\\
= a{b^2}{c^2}\left( {b - a} \right) + ab{c^3}\left( {b - a} \right)\\
= \left( {b - a} \right)\left( {a{b^2}{c^2} + ab{c^3}} \right)\\
= \left( {b - a} \right).ab{c^2}.\left( {b + c} \right)\\
= ab{c^2}\left( {b - a} \right)\left( {b + c} \right)\\
b,\\
B = 7 - 12x - 4{x^2}\\
= \left( { - 4{x^2} + 2x} \right) - \left( {14x - 7} \right)\\
= - 2x\left( {2x - 1} \right) - 7\left( {2x - 1} \right)\\
= - \left( {2x - 1} \right)\left( {2x + 7} \right)\\
c,\\
C = 4{x^2} - 15x + 14\\
= \left( {4{x^2} - 8x} \right) - \left( {7x - 14} \right)\\
= 4x\left( {x - 2} \right) - 7\left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {4x - 7} \right)\\
d,\\
D = 5{x^2} - 30xy + 25{y^2}\\
= 5\left( {{x^2} - 6xy + 5{y^2}} \right)\\
= 5.\left[ {\left( {{x^2} - xy} \right) - \left( {5xy - 5{y^2}} \right)} \right]\\
= 5.\left[ {x\left( {x - y} \right) - 5y\left( {x - y} \right)} \right]\\
= 5.\left( {x - y} \right)\left( {x - 5y} \right)
\end{array}\)
