Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
{a^2} + {b^2} + 2ab - 1\\
= \left( {{a^2} + 2ab + {b^2}} \right) - 1\\
= {\left( {a + b} \right)^2} - {1^2}\\
= \left( {a + b - 1} \right)\left( {a + b + 1} \right)\\
c,\\
{a^2} - 2a + 1 - 9{b^2}\\
= \left( {{a^2} - 2a + 1} \right) - 9{b^2}\\
= {\left( {a - 1} \right)^2} - {\left( {3b} \right)^2}\\
= \left( {a - 1 - 3b} \right)\left( {a - 1 + 3b} \right)\\
d,\\
{x^2} - 6xy + 9{y^2} - 36\\
= \left( {{x^2} - 6xy + 9{y^2}} \right) - 36\\
= {\left( {x - 3y} \right)^2} - {6^2}\\
= \left( {x - 3y - 6} \right)\left( {x - 3y + 6} \right)\\
e,\\
16 - {x^2} + 2xy - {y^2}\\
= 16 - \left( {{x^2} - 2xy + {y^2}} \right)\\
= {4^2} - {\left( {x - y} \right)^2}\\
= \left[ {4 - \left( {x - y} \right)} \right].\left[ {4 + \left( {x - y} \right)} \right]\\
= \left( { - x + y + 4} \right)\left( {x - y + 4} \right)\\
f,\\
4{y^2} - 4{x^2} - 4y + 1\\
= \left( {4{y^2} - 4y + 1} \right) - 4{x^2}\\
= {\left( {2y - 1} \right)^2} - {\left( {2x} \right)^2}\\
= \left( {2y - 1 - 2x} \right)\left( {2y - 1 + 2x} \right)
\end{array}\)
Đề câu a sai hay sao rồi ấy em??
