Giải thích các bước giải:
a) Ta có;
$\begin{array}{l}
A = {x^3} - 30{x^2} - 31x + 1\\
= \left( {{x^3} - 31{x^2}} \right) + \left( {{x^2} - 31x} \right) + 1\\
= {x^2}\left( {x - 31} \right) + x\left( {x - 31} \right) + 1
\end{array}$
Khi $x = 31 \Rightarrow x - 31 = 0 \Rightarrow A = 1$
Vậy $A=1$ khi $x=31$
b) Ta có:
$\begin{array}{l}
B = {x^5} - 15{x^4} + 16{x^3} - 29{x^2} + 13x\\
= \left( {{x^5} - 14{x^4}} \right) - \left( {{x^4} - 14{x^3}} \right) + 2\left( {{x^3} - 14{x^2}} \right) - \left( {{x^2} - 14x} \right) - x\\
= {x^4}\left( {x - 14} \right) - {x^3}\left( {x - 14} \right) + 2{x^2}\left( {x - 14} \right) - x\left( {x - 14} \right) - x
\end{array}$
Khi $x = 14 \Rightarrow x - 14 = 0 \Rightarrow B = - 14$
Vậy $B=-14$
c) Ta có:
Khi $x=9$ thì:
$\begin{array}{l}
C = {x^{14}} - 10{x^{13}} + 10{x^{12}} - 10{x^{11}} + ... + 10{x^2} - 10x + 10\\
= {x^{14}} - \left( {x + 1} \right){x^{13}} + \left( {x + 1} \right){x^{12}} - \left( {x + 1} \right){x^{11}} + ... + \left( {x + 1} \right){x^2} - \left( {x + 1} \right)x + x + 1\\
= {x^{14}} - {x^{14}} - {x^{13}} + {x^{13}} + {x^{12}} - {x^{12}} - {x^{11}} + ... + {x^3} + {x^2} - {x^2} - x + x + 1\\
= 1
\end{array}$
