Ta có $Q(x)\quad\vdots\quad 3, \forall x\in Z$
$\to \begin{cases} Q(0)\quad\vdots\quad 3\\ Q(1)\quad\vdots\quad 3\\ Q(-1)\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}a\cdot 0^3+b\cdot 0^2+c\cdot 0+d\quad\vdots\quad 3\\ a\cdot 1^3+b\cdot 1^2+c\cdot 1+d\quad\vdots\quad 3\\ a\cdot (-1)^3+b\cdot (-1)^2+c\cdot (-1)+d\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+b+c+d\quad\vdots\quad 3\\ -a+b-c+d\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+b+c\quad\vdots\quad 3\\ -a+b-c\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+b+c\quad\vdots\quad 3\\ (-a+b-c)+(a+b+c)\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+b+c\quad\vdots\quad 3\\ 2b\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+b+c\quad\vdots\quad 3\\ b\quad\vdots\quad 3\end{cases}$
$\to \begin{cases}d\quad\vdots\quad 3\\ a+c\quad\vdots\quad 3\\ b\quad\vdots\quad 3\end{cases}$
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dạng giống nha bn
Ta có: \(Q\left(x\right)⋮3\) ; \(\forall x\)
\(\Rightarrow Q\left(0\right)=a.0^3+b.0^2+c.0+d=d\)
\(\Leftrightarrow d⋮3\)
Xét \(Q\left(1\right)⋮3\)
\(\Rightarrow a+b+c+d⋮3\)
Mà \(d⋮3\) \(\Rightarrow a+b+c⋮3\) (1)
Xét \(Q\left(-1\right)⋮3\)
\(\Rightarrow-a+b-c+d⋮3\)
Mà \(d⋮3\) \(\Rightarrow-a+b-c⋮3\) (2)
\(\left(1\right);\left(2\right)\Rightarrow\left(a+b+c\right)+\left(-a+b-c\right)⋮3\)
\(\Rightarrow2b⋮3\)
mà UCLN \(\left(2;3\right)=1\) \(\Rightarrow b⋮3\)
Ta có: \(a+b+c⋮3\)
\(\Rightarrow a+c⋮3\) (3)
Xét \(Q\left(2\right)⋮3\)
\(\Rightarrow8a+4b+2c+d⋮3\)
Mà \(d⋮3;4b⋮3\) ( vì \(b⋮3\) )
\(\Rightarrow8a+2c⋮3\)
\(\Leftrightarrow2\left(4a+c\right)⋮3\)
Mà UCLN \(\left(2;3\right)=1\)
\(\Rightarrow4a+c⋮3\) (4)
\(\left(3\right);\left(4\right)\Rightarrow\left(4a+c\right)-\left(a+c\right)⋮3\)
\(\Leftrightarrow3a⋮3\)
\(\Rightarrow a⋮3\)
Mà \(a+c⋮3\) \(\Rightarrow c⋮3\)
Vậy \(a,b,c,d⋮3\) ;\(\forall x\)
