Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
4,\\
A = 2{x^3} - {x^2} - x + 1\\
 = \left( {2{x^3} - 4{x^2}} \right) + \left( {3{x^2} - 6x} \right) + 5x + 1\\
 = 2x.\left( {{x^2} - 2x} \right) + 3.\left( {{x^2} - 2x} \right) + 5x + 1\\
 = \left( {{x^2} - 2x} \right)\left( {2x + 3} \right) + 5x + 1\\
 = B.\left( {2x + 3} \right) + \left( {5x + 1} \right)\\
 \Rightarrow Q = 2x + 3\\
 \Rightarrow R = 5x + 1\\
5,\\
{\left( {2n - 1} \right)^3} - \left( {2n - 1} \right)\\
 = \left( {2n - 1} \right).\left[ {{{\left( {2n - 1} \right)}^2} - 1} \right]\\
 = \left( {2n - 1} \right).\left[ {\left( {2n - 1} \right) - 1} \right].\left[ {\left( {2n - 1} \right) + 1} \right]\\
 = \left( {2n - 1} \right).\left( {2n - 2} \right).2n\\
 = \left( {2n - 1} \right).2.\left( {n - 1} \right).2n\\
 = 4n\left( {n - 1} \right)\left( {2n - 1} \right)
\end{array}\)

\(n - 1;\,\,n\) là 2 số nguyên liên tiếp nên \(\left( {n - 1} \right)n\,\, \vdots \,\,2\)

Suy ra \(4n\left( {n - 1} \right)\left( {2n - 1} \right)\,\, \vdots \,\,8\)

\(\begin{array}{l}
6,\\
P = {\left( {x - y} \right)^2} + {\left( {x + y} \right)^2} - 2.\left( {x + y} \right)\left( {x - y} \right) - 4{x^2}\\
 = \left[ {{{\left( {x + y} \right)}^2} - 2.\left( {x + y} \right).\left( {x - y} \right) + {{\left( {x - y} \right)}^2}} \right] - 4{x^2}\\
 = {\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right]^2} - 4{x^2}\\
 = {\left( {2y} \right)^2} - 4{x^2}\\
 = 4{y^2} - 4{x^2}\\
Q = 2.\left( {x + y} \right)\left( {x - y} \right) - {\left( {x - y} \right)^2} + {\left( {x + y} \right)^2} - 4{y^2}\\
 = 2.\left( {{x^2} - {y^2}} \right) - \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{x^2} + 2xy + {y^2}} \right) - 4{y^2}\\
 = 2{x^2} - 2{y^2} - {x^2} + 2xy - {y^2} + {x^2} + 2xy + {y^2} - 4{y^2}\\
 = 2{x^2} + 4xy - 6{y^2}
\end{array}\)