Đáp án:
\(\begin{array}{l}
g,\\
{\left( {a - 1} \right)^2}.{\left( {a + 1} \right)^2}\\
h,\\
\left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
i,\\
\left( {x - y - 2} \right)\left( {x + y - 2} \right)\\
k,\\
\left( {x - 1} \right)\left( {{x^3} - x - 1} \right)\\
l,\\
3x\left( {x + 2} \right)\\
m,\\
\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
{\left( {{a^2} + 1} \right)^2} - 4{a^2}\\
= {\left( {{a^2} + 1} \right)^2} - {\left( {2a} \right)^2}\\
= \left[ {\left( {{a^2} + 1} \right) - 2a} \right].\left[ {\left( {{a^2} + 1} \right) + 2a} \right]\\
= \left( {{a^2} - 2a + 1} \right)\left( {{a^2} + 2a + 1} \right)\\
= \left( {{a^2} - 2.a.1 + {1^2}} \right).\left( {{a^2} + 2.a.1 + {1^2}} \right)\\
= {\left( {a - 1} \right)^2}.{\left( {a + 1} \right)^2}\\
h,\\
{x^3} - 3{x^2} - 4x + 12\\
= \left( {{x^3} - 3{x^2}} \right) + \left( { - 4x + 12} \right)\\
= {x^2}.\left( {x - 3} \right) - 4.\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - {2^2}} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
i,\\
{x^2} - {y^2} + 4 - 4x\\
= \left( {{x^2} - 4x + 4} \right) - {y^2}\\
= \left( {{x^2} - 2.x.2 + {2^2}} \right) - {y^2}\\
= {\left( {x - 2} \right)^2} - {y^2}\\
= \left[ {\left( {x - 2} \right) - y} \right].\left[ {\left( {x - 2} \right) + y} \right]\\
= \left( {x - y - 2} \right)\left( {x + y - 2} \right)\\
k,\\
{x^4} - {x^3} - {x^2} + 1\\
= \left( {{x^4} - {x^3}} \right) - \left( {{x^2} - 1} \right)\\
= {x^3}.\left( {x - 1} \right) - \left( {{x^2} - {1^2}} \right)\\
= {x^3}\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right)\\
= \left( {x - 1} \right).\left[ {{x^3} - \left( {x + 1} \right)} \right]\\
= \left( {x - 1} \right)\left( {{x^3} - x - 1} \right)\\
l,\\
{\left( {2x + 1} \right)^2} - {\left( {x - 1} \right)^2}\\
= \left[ {\left( {2x + 1} \right) - \left( {x - 1} \right)} \right].\left[ {\left( {2x + 1} \right) + \left( {x - 1} \right)} \right]\\
= \left( {2x + 1 - x + 1} \right)\left( {2x + 1 + x - 1} \right)\\
= \left( {x + 2} \right).3x\\
= 3x\left( {x + 2} \right)\\
m,\\
{x^4} + 4{x^2} - 5\\
= \left( {{x^4} + 4{x^2} + 4} \right) - 9\\
= \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.2 + {2^2}} \right] - {3^2}\\
= {\left( {{x^2} + 2} \right)^2} - {3^2}\\
= \left[ {\left( {{x^2} + 2} \right) - 3} \right].\left[ {\left( {{x^2} + 2} \right) + 3} \right]\\
= \left( {{x^2} - 1} \right)\left( {{x^2} + 5} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)
\end{array}\)
