1)
$\begin{array}{l}
{n^3} - 3{n^2} - 3n - 1 = \\
= {n^3} + {n^2} + n - = {n^3} + {n^2} + n - 4{n^2} - 4n - 4 + 3\\
= \left( {{n^2} + n + 1} \right)\left( {n - 4} \right) + 3 \vdots {n^2} + n + 1\\
\Rightarrow 3 \vdots {n^2} + n + 1,{n^2} + n + 1 > 0\\
{n^2} + n + 1 = \left\{ {1,3} \right\}\\
T{H_1}:{n^2} + n + 1 = 1\\
\Rightarrow n\left( {n + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
n = 0\\
n + 1 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
n = 0\\
n = - 1
\end{array} \right.\\
T{H_1}:{n^2} + n + 1 = 3\\
{n^2} + n - 2 = 0 \Rightarrow \left( {n + 2} \right)\left( {n - 1} \right)\\
\Rightarrow \left[ \begin{array}{l}
n + 2 = 0\\
n - 1 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
n = - 2\\
n = 1
\end{array} \right.
\end{array}$
b)
$\begin{array}{l}
103{n^2} + 121n + 70 \vdots n - 1\\
103{n^2} - 103n + 224n - 224 + 294 \vdots n - 1\\
\Rightarrow 294 \vdots n - 1
\end{array}$
tương tự câu 1
c)
$\begin{array}{l}
{n^3} - {n^2} + 2n + 7 \vdots {n^2} + 1\\
= {n^3} + n - {n^2} - 1 + n + 8 \vdots {n^2} + 1\\
\Rightarrow n + 8 \vdots {n^2} + 1\\
\Rightarrow \left( {n + 8} \right)\left( {n - 8} \right) \vdots {n^2} + 1\\
\Rightarrow {n^2} + 1 - 65 \vdots {n^2} + 1\\
\Rightarrow 65 \vdots {n^2} + 1
\end{array}$
tương tự câu 1 và 2
2)
\begin{array}{l}
1 + x + {x^{19}} + {x^{199}} + {x^{1995}}\\
${x^{1995}} - x = x\left( {{x^{1994}} - 1} \right) = x\left( {{x^{1994}} - {1^{998}}} \right) \vdots 1 - {x^2}\\
{x^{199}} - x = x\left( {{x^{198}} - 1} \right) = x\left( {{x^{198}} - {1^{99}}} \right) \vdots 1 - {x^2}\\
{x^{19}} - x = x\left( {{x^{18}} - 1} \right) = x\left( {{x^{18}} - {1^9}} \right) \vdots 1 - {x^2}\\
\Rightarrow 1 + x + {x^{19}} + {x^{19}} + {x^{1995}} = {x^{1995}} - x + {x^{199}} - x + {x^{19}} - x + 4x + 1\\
\Rightarrow 1 + x + {x^{19}} + {x^{19}} + {x^{1995}}:1 - {x^2} dư 4x+1
\end{array}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
