Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{a^2} - 4{b^2} + 2a + 4b\\
= \left( {{a^2} - {{\left( {2b} \right)}^2}} \right) + \left( {2a + 4b} \right)\\
= \left( {a - 2b} \right)\left( {a + 2b} \right) + 2.\left( {a + 2b} \right)\\
= \left( {a + 2b} \right).\left( {a - 2b + 2} \right)\\
b,\\
{x^2} + 16{y^2} - 8xy - 36\\
= \left( {{x^2} - 8xy + 16{y^2}} \right) - 36\\
= \left[ {{x^2} - 2.x.4y + {{\left( {4y} \right)}^2}} \right] - {6^2}\\
= {\left( {x - 4y} \right)^2} - {6^2}\\
= \left( {x - 4y + 6} \right)\left( {x - 4y - 6} \right)\\
3,\\
a,\\
7{x^2} + 9x - 5 = x.\left( {2 + 7x} \right)\\
\Leftrightarrow 7{x^2} + 9x - 5 = 2x + 7{x^2}\\
\Leftrightarrow 9x - 2x = 5\\
\Leftrightarrow 7x = 5\\
\Leftrightarrow x = \frac{5}{7}\\
b,\\
{x^2} - 5x - 3x + 15 = 0\\
\Leftrightarrow \left( {{x^2} - 5x} \right) - \left( {3x - 15} \right) = 0\\
\Leftrightarrow x.\left( {x - 5} \right) - 3.\left( {x - 5} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.
\end{array}\)
