Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^4} + 2{x^3} + {x^2}\\
= {x^2}\left( {{x^2} + 2x + 1} \right)\\
= {x^2}.{\left( {x + 1} \right)^2}\\
b,\\
{x^4} - 3{x^3} - x + 3\\
= \left( {{x^4} - 3{x^2}} \right) - \left( {x - 3} \right)\\
= {x^3}\left( {x - 3} \right) - \left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^3} - 1} \right)\\
= \left( {x - 3} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
c,\\
{x^3} - {x^2}y - x{y^2} + {y^3}\\
= \left( {{x^3} - {x^2}y} \right) - \left( {x{y^2} - {y^3}} \right)\\
= {x^2}\left( {x - y} \right) - {y^2}\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {{x^2} - {y^2}} \right)\\
= \left( {x - y} \right)\left( {x - y} \right)\left( {x + y} \right)\\
= {\left( {x - y} \right)^2}\left( {x + y} \right)\\
d,\\
3x + 3y - {x^2} - 2xy - {y^2}\\
= \left( {3x + 3y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
= 3.\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
= \left( {x + y} \right)\left( {3 - x - y} \right)\\
e,\\
4{x^2} + 4x - 3\\
= \left( {4{x^2} - 2x} \right) + \left( {6x - 3} \right)\\
= 2x.\left( {2x - 1} \right) + 3.\left( {2x - 1} \right)\\
= \left( {2x - 1} \right)\left( {2x + 3} \right)\\
h,\\
{x^4} + 4{y^4}\\
= \left( {{x^4} + 4{x^2}{y^2} + 4{y^4}} \right) - 4{x^2}{y^2}\\
= {\left( {{x^2} + 2{y^2}} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {{x^2} + 2{y^2} - 2xy} \right)\left( {{x^2} + 2{y^2} + 2xy} \right)\\
k,\\
{x^5} + {x^4} + 1\\
= \left( {{x^5} + {x^4} + {x^3}} \right) - \left( {{x^3} + {x^2} + x} \right) + \left( {{x^2} + x + 1} \right)\\
= {x^3}\left( {{x^2} + x + 1} \right) - x\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^3} - x + 1} \right)
\end{array}\)
Em xem lại đề câu g nhé!
