Đáp án:
\(\begin{array}{l}
a,\\
5y.\left( {5y + 3} \right)\\
b,\\
\left( {x - y + 1} \right)\left( {x + y + 1} \right)\\
c,\\
\left( {x - y} \right)\left( {x + y + 3} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
25{y^2} + 15y\\
= 5y.5y + 3.5y\\
= 5y.\left( {5y + 3} \right)\\
b,\\
{x^2} + 2x - {y^2} + 1\\
= \left( {{x^2} + 2x + 1} \right) - {y^2}\\
= \left( {{x^2} + 2.x.1 + {1^2}} \right) - {y^2}\\
= {\left( {x + 1} \right)^2} - {y^2}\\
= \left[ {\left( {x + 1} \right) - y} \right].\left[ {\left( {x + 1} \right) + y} \right]\\
= \left( {x - y + 1} \right)\left( {x + y + 1} \right)\\
c,\\
3x - 3y + {x^2} - {y^2}\\
= \left( {3x - 3y} \right) + \left( {{x^2} - {y^2}} \right)\\
= 3.\left( {x - y} \right) + \left( {x - y} \right)\left( {x + y} \right)\\
= \left( {x - y} \right).\left[ {3 + \left( {x + y} \right)} \right]\\
= \left( {x - y} \right)\left( {x + y + 3} \right)
\end{array}\)
