Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x - 2y + {x^2} - 4{y^2}\\
= \left( {x - 2y} \right) + \left( {{x^2} - 4{y^2}} \right)\\
= \left( {x - 2y} \right) + \left( {x - 2y} \right)\left( {x + 2y} \right)\\
= \left( {x - 2y} \right)\left( {1 + x + 2y} \right)\\
c,\\
{x^6} - {x^4} + 2{x^3} + 2{x^2}\\
= {x^2}.\left( {{x^4} - {x^2} + 2x + 2} \right)\\
= {x^2}.\left[ {\left( {{x^4} - {x^2}} \right) + \left( {2x + 2} \right)} \right]\\
= {x^2}.\left[ {{x^2}\left( {{x^2} - 1} \right) + 2\left( {x + 1} \right)} \right]\\
= {x^2}.\left[ {{x^2}\left( {x - 1} \right)\left( {x + 1} \right) + 2\left( {x + 1} \right)} \right]\\
= {x^2}.\left( {x + 1} \right).\left[ {{x^2}\left( {x - 1} \right) + 2} \right]\\
= {x^2}\left( {x + 1} \right)\left( {{x^3} - {x^2} + 2} \right)\\
= {x^2}\left( {x + 1} \right).\left[ {\left( {{x^3} + {x^2}} \right) - \left( {2{x^2} + 2x} \right) + \left( {2x + 2} \right)} \right]\\
= {x^2}\left( {x + 1} \right).\left[ {{x^2}\left( {x + 1} \right) - 2x\left( {x + 1} \right) + 2\left( {x + 1} \right)} \right]\\
= {x^2}.{\left( {x + 1} \right)^2}.\left( {{x^2} - 2x + 2} \right)\\
b,\\
{x^2} - 4{x^2}{y^2} + {y^2} + 2xy\\
= \left( {{x^2} + 2xy + {y^2}} \right) - 4{x^2}{y^2}\\
= {\left( {x + y} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {x + y + 2xy} \right)\left( {x + y - 2xy} \right)\\
d,\\
{x^3} + 3{x^2} + 3x + 1 - 8{y^3}\\
= \left( {{x^3} + 3{x^2} + 3x + 1} \right) - 8{y^3}\\
= {\left( {x + 1} \right)^3} - {\left( {2y} \right)^3}\\
= \left( {x + 1 - 2y} \right)\left[ {{{\left( {x + 1} \right)}^2} - \left( {x + 1} \right).2y + 4{y^2}} \right]\\
b,\\
{\left( {3x - 1} \right)^2} - 9 = 0\\
\Leftrightarrow {\left( {3x - 1} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 3\\
3x - 1 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{2}{3}
\end{array} \right.\\
a,\\
{x^2} - 4x + 3 = \left( {{x^2} - x} \right) - \left( {3x - 3} \right) = x\left( {x - 1} \right) - 3\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 3} \right)\\
c,\\
2{x^2} - 3x - 2 = \left( {2{x^2} - 4x} \right) + \left( {x - 2} \right) = 2x\left( {x - 2} \right) + \left( {x - 2} \right) = \left( {x - 2} \right)\left( {2x + 1} \right)\\
b,\\
{x^2} + 4x - 12 = \left( {{x^2} + 6x} \right) - \left( {2x + 12} \right) = x\left( {x + 6} \right) - 2\left( {x + 6} \right) = \left( {x + 6} \right)\left( {x - 2} \right)\\
d,\\
2{x^3} + x - 2{x^2} = x\left( {2{x^2} - 2x + 1} \right)
\end{array}\)
