Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {2{x^4} - 5{x^2} + {x^3} - 3 - 3x} \right):\left( {{x^2} - 3} \right)\\
= \left[ {\left( {2{x^4} - 6{x^2}} \right) + \left( {{x^2} - 3} \right) + \left( {{x^3} - 3x} \right)} \right]:\left( {{x^2} - 3} \right)\\
= \left[ {2{x^2}\left( {{x^2} - 3} \right) + \left( {{x^2} - 3} \right) + x\left( {{x^2} - 3} \right)} \right]:\left( {{x^2} - 3} \right)\\
= \left[ {\left( {{x^2} - 3} \right).\left( {2{x^2} + x + 1} \right)} \right]:\left( {{x^2} - 3} \right)\\
= 2{x^2} + x + 1\\
b,\\
\left( {{x^5} + {x^3} + {x^2} + 1} \right):\left( {{x^3} + 1} \right)\\
= \left[ {\left( {{x^5} + {x^2}} \right) + \left( {{x^3} + 1} \right)} \right]:\left( {{x^3} + 1} \right)\\
= \left[ {{x^2}\left( {{x^3} + 1} \right) + \left( {{x^3} + 1} \right)} \right]:\left( {{x^3} + 1} \right)\\
= \left[ {\left( {{x^3} + 1} \right)\left( {{x^2} + 1} \right)} \right]:\left( {{x^3} + 1} \right)\\
= {x^2} + 1\\
c,\\
\left( {2{x^3} + 5{x^2} - 2x + 3} \right):\left( {2{x^2} - x + 1} \right)\\
= \left[ {\left( {2{x^3} - {x^2} + x} \right) + \left( {6{x^2} - 3x + 3} \right)} \right]:\left( {2{x^2} - x + 1} \right)\\
= \left[ {x\left( {2{x^2} - x + 1} \right) + 3.\left( {2{x^2} - x + 1} \right)} \right]:\left( {2{x^2} - x + 1} \right)\\
= \left[ {\left( {2{x^2} - x + 1} \right)\left( {x + 3} \right)} \right]:\left( {2{x^2} - x + 1} \right)\\
= x + 3\\
d,\\
\left( { - {x^3} + 2{x^4} - 4 - {x^2} + 7x} \right):\left( {{x^2} + x - 1} \right)\\
= \left[ {\left( {2{x^4} + 2{x^3} - 2{x^2}} \right) - \left( {3{x^3} + 3{x^2} - 3x} \right) + \left( {4{x^2} + 4x - 4} \right)} \right]:\left( {{x^2} + x - 1} \right)\\
= \left[ {2{x^2}\left( {{x^2} + x - 1} \right) - 3x\left( {{x^2} + x - 1} \right) + 4\left( {{x^2} + x - 1} \right)} \right]:\left( {{x^2} + x - 1} \right)\\
= \left[ {\left( {{x^2} + x - 1} \right)\left( {2{x^2} - 3x + 4} \right)} \right]:\left( {{x^2} + x - 1} \right)\\
= 2{x^2} - 3x + 4
\end{array}\)
