Đáp án:
\(\begin{array}{l}
a,\\
x.\left( {x + 1} \right).\left( {{x^2} - x + 2} \right)\\
b,\\
\left( {9{x^2} - 6x + 2} \right).\left( {9{x^2} + 6x + 2} \right)\\
c,\\
\left( {{x^2} - x + 1} \right).\left( {{x^2} + x + 1} \right)\\
d,\\
\left( {{x^2}{y^2} - 2xy + 2} \right).\left( {{x^2}{y^2} + 2xy + 2} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^4} + {x^2} + 2x\\
= x.\left( {{x^3} + x + 2} \right)\\
= x.\left[ {\left( {{x^3} + {x^2}} \right) + \left( { - {x^2} - x} \right) + \left( {2x + 2} \right)} \right]\\
= x.\left[ {{x^2}\left( {x + 1} \right) - x.\left( {x + 1} \right) + 2.\left( {x + 1} \right)} \right]\\
= x.\left( {x + 1} \right).\left( {{x^2} - x + 2} \right)\\
b,\\
81{x^4} + 4\\
= \left( {81{x^4} + 36{x^2} + 4} \right) - 36{x^2}\\
= \left[ {{{\left( {9{x^2}} \right)}^2} + 2.9{x^2}.2 + {2^2}} \right] - {\left( {6x} \right)^2}\\
= {\left( {9{x^2} + 2} \right)^2} - {\left( {6x} \right)^2}\\
= \left[ {\left( {9{x^2} + 2} \right) - 6x} \right].\left[ {\left( {9{x^2} + 2} \right) + 6x} \right]\\
= \left( {9{x^2} - 6x + 2} \right).\left( {9{x^2} + 6x + 2} \right)\\
c,\\
{x^4} + {x^2} + 1\\
= \left( {{x^4} + 2{x^2} + 1} \right) - {x^2}\\
= \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.1 + {1^2}} \right] - {x^2}\\
= {\left( {{x^2} + 1} \right)^2} - {x^2}\\
= \left[ {\left( {{x^2} + 1} \right) - x} \right].\left[ {\left( {{x^2} + 1} \right) + x} \right]\\
= \left( {{x^2} - x + 1} \right).\left( {{x^2} + x + 1} \right)\\
d,\\
{x^4}{y^4} + 4\\
= \left( {{x^4}{y^4} + 4{x^2}{y^2} + 4} \right) - 4{x^2}{y^2}\\
= \left[ {{{\left( {{x^2}{y^2}} \right)}^2} + 2.{x^2}{y^2}.2 + {2^2}} \right] - {\left( {2xy} \right)^2}\\
= {\left( {{x^2}{y^2} + 2} \right)^2} - {\left( {2xy} \right)^2}\\
= \left[ {\left( {{x^2}{y^2} + 2} \right) - 2xy} \right].\left[ {\left( {{x^2}{y^2} + 2} \right) + 2xy} \right]\\
= \left( {{x^2}{y^2} - 2xy + 2} \right).\left( {{x^2}{y^2} + 2xy + 2} \right)
\end{array}\)
