Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {x + 5} \right)^2} - {\left( {3 - x} \right)^2}\\
= \left[ {\left( {x + 5} \right) - \left( {3 - x} \right)} \right].\left[ {\left( {x + 5} \right) + \left( {3 - x} \right)} \right]\\
= \left( {2x + 2} \right).8\\
= 16x + 16\\
b,\\
{\left( {4 - x} \right)^2} - {\left( {x - 3} \right)^2}\\
= \left[ {\left( {4 - x} \right) - \left( {x - 3} \right)} \right].\left[ {\left( {4 - x} \right) + \left( {x - 3} \right)} \right]\\
= \left( {7 - 2x} \right).1\\
= 7 - 2x\\
c,\\
\left( {x - 5} \right)\left( {x + 5} \right) - {\left( {x + 5} \right)^2}\\
= \left( {x + 5} \right).\left[ {\left( {x - 5} \right) - \left( {x + 5} \right)} \right]\\
= \left( {x + 5} \right).\left( { - 10} \right)\\
= - 10x - 50\\
d,\\
{\left( {x - 3} \right)^2} - \left( {x - 4} \right)\left( {x + 4} \right)\\
= \left( {{x^2} - 6x + 9} \right) - \left( {{x^2} - 16} \right)\\
= {x^2} - 6x + 9 - {x^2} + 16\\
= - 6x + 25
\end{array}\)
