Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
4{x^2} - 25 - \left( {2x + 7} \right)\left( {5 - 2x} \right)\\
 = {\left( {2x} \right)^2} - {5^2} + \left( {2x + 7} \right)\left( {2x - 5} \right)\\
 = \left( {2x - 5} \right)\left( {2x + 5} \right) + \left( {2x + 7} \right)\left( {2x - 5} \right)\\
 = \left( {2x - 5} \right).\left[ {\left( {2x + 5} \right) + \left( {2x + 7} \right)} \right]\\
 = \left( {2x - 5} \right)\left( {4x + 12} \right)\\
b,\\
4{x^2} - 9{y^2} - 10x + 15y\\
 = \left( {4{x^2} - 9{y^2}} \right) + \left( { - 10x + 15y} \right)\\
 = \left[ {{{\left( {2x} \right)}^2} - {{\left( {3y} \right)}^2}} \right] - 5.\left( {2x - 3y} \right)\\
 = \left( {2x - 3y} \right)\left( {2x + 3y} \right) - 5.\left( {2x - 3y} \right)\\
 = \left( {2x - 3y} \right)\left( {2x + 3y - 5} \right)\\
c,\\
{\left( {ax + by} \right)^2} - {\left( {ax - by} \right)^2}\\
 = \left[ {\left( {ax + by} \right) - \left( {ax - by} \right)} \right].\left[ {\left( {ax + by} \right) + \left( {ax - by} \right)} \right]\\
 = 2by.2ax\\
 = 4axby\\
e,\\
{x^2} - 10x + 16\\
 = \left( {{x^2} - 2x} \right) + \left( { - 8x + 16} \right)\\
 = \left( {{x^2} - 2x} \right) - \left( {8x - 16} \right)\\
 = x\left( {x - 2} \right) - 8.\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x - 8} \right)\\
f,\\
{a^2}\left( {b - c} \right) + {b^2}\left( {c - a} \right) + {c^2}\left( {a - b} \right)\\
 = {a^2}\left( {b - c} \right) + {b^2}c - {b^2}a + {c^2}a - {c^2}b\\
 = {a^2}\left( {b - c} \right) + \left( {{b^2}c - {c^2}b} \right) + \left( {{c^2}a - {b^2}a} \right)\\
 = {a^2}\left( {b - c} \right) + bc.\left( {b - c} \right) + a.\left( {{c^2} - {b^2}} \right)\\
 = {a^2}\left( {b - c} \right) + bc\left( {b - c} \right) + a.\left( {c + b} \right)\left( {c - b} \right)\\
 = \left( {b - c} \right).\left( {{a^2} + bc - a.\left( {c + b} \right)} \right)\\
 = \left( {b - c} \right).\left( {{a^2} + bc - ac - ab} \right)\\
 = \left( {b - c} \right).\left[ {\left( {{a^2} - ac} \right) + \left( {bc - ab} \right)} \right]\\
 = \left( {b - c} \right).\left[ {a.\left( {a - c} \right) + b\left( {c - a} \right)} \right]\\
 = \left( {b - c} \right).\left( {a - c} \right)\left( {a - b} \right)\\
g,\\
{\left( {{a^2} + {b^2} - 5} \right)^2} - 4.{\left( {ab + 2} \right)^2}\\
 = {\left( {{a^2} + {b^2} - 5} \right)^2} - {\left[ {2.\left( {ab + 2} \right)} \right]^2}\\
 = {\left( {{a^2} + {b^2} - 5} \right)^2} - {\left( {2ab + 4} \right)^2}\\
 = \left[ {\left( {{a^2} + {b^2} - 5} \right) - \left( {2ab + 4} \right)} \right].\left[ {\left( {{a^2} + {b^2} - 5} \right) + \left( {2ab + 4} \right)} \right]\\
 = \left[ {{a^2} + {b^2} - 2ab - 9} \right].\left[ {{a^2} + {b^2} + 2ab - 1} \right]\\
 = \left[ {\left( {{a^2} - 2ab + {b^2}} \right) - 9} \right].\left[ {\left( {{a^2} + 2ab + {b^2}} \right) - 1} \right]\\
 = \left[ {{{\left( {a - b} \right)}^2} - 9} \right].\left[ {{{\left( {a + b} \right)}^2} - 1} \right]\\
 = \left( {a - b - 3} \right)\left( {a - b + 3} \right).\left( {a + b - 1} \right)\left( {a + b + 1} \right)
\end{array}\)