Question

Cho đa thức f (x) = ax² + bx + c ( a,b,c \(\in\) Z ) Biết f (-1) ; f (0) ; f (1) đều chia hết cho 3 . CM : a\(⋮\) 3; b \(⋮\) 3; c\(⋮\) 3

Answer 1

Ta có: f(0) = c \(⋮\) 3

f(1) = a + b + c \(⋮\) 3 \(\Rightarrow\) a + b \(⋮\) 3 (1)

f(-1) = a - b + c \(⋮\) 3 \(\Rightarrow\) a - b \(⋮\) 3 (2)

Từ (1) và (2) suy ra a + b + a - b \(⋮\) 3 và a + b - a + b \(⋮\) 3

\(\Rightarrow\) \(\left\{{}\begin{matrix}2a⋮3\\2b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\end{matrix}\right.\)

Vậy a, b, c \(⋮\) 3

Answer 2

+ \(\left\{{}\begin{matrix}f\left(0\right)⋮3\\f\left(1\right)⋮3\\f\left(-1\right)⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c⋮3\\a+b+c⋮3\\a-b+c⋮3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b⋮3\\a-b⋮3\\c⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a⋮3\\-2b⋮3\\c⋮3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\\c⋮3\end{matrix}\right.\)