Ta có:

\(f\left(5\right)=125a+25b+5c+d\)

\(f\left(4\right)=64a+16b+4c+d\)

\(f\left(7\right)=343a+49b+7c+d\)

\(f\left(2\right)=8a+4b+2c+d\)

Xét:

\(f\left(5\right)-f\left(4\right)=125a+25b+5c+d-64a-16b-4c-d\)

\(=61a+9b+c=2019\)

Khi đó:

\(f\left(7\right)-f\left(2\right)=343a+49b+7c+d-8a-4b-2c-d\)

\(=335a+45b+5c=5\left(61a+9b+c\right)+30=5\cdot2019+30⋮5\)

Vậy ta có đpcm

không ra được đâu, 335 không chia hết cho 61, 5.61=305 chứ không phải bằng 335

* Ta có A(x)=ax^3+bx^2+cx+d

=>A(5)=125a+25b+5c+d

    A(4)=64a+16b+4c+d

    A(7)=343a+49b+7c+d

    A(2)=8a+4b+2c+d

+)Có A(5)-A(4)=(125a+25b+5c+d)-(64a+16b+4c+d)

    =>A(5)-A(4)=61a+9b+c

+) Xét A(7)-A(2)=(343a+49b+7c+d)-(8a+4b+2c+d)

      =>A(7)-A(2)=335a+45b+5c

                         =(61a+9b+c).5+30a

                         =(2022.5+30a) chia hết cho 2

         Vì a thuộc Z+ nên 2022.5+30a>2 nên A(7)-A(2) là hợp số

Giải:

Ta có: \(f\left(5\right)-f\left(4\right)=2012\)

\(\Leftrightarrow\left(125a+25b+5c+d\right)\)\(-\left(64a+16b+4c+d\right)=2012\)

\(\Leftrightarrow61a+9b+c=2012\)

Lại có: \(f\left(7\right)-f\left(2\right)\)

\(=\left(343a+49b+7c+d\right)-\) \(\left(8a+4b+2c+d\right)\)

\(=335a+45b+5c=305a+45b+5c+30a\)

\(=5\left(61a+9b+c\right)+30a=2012+30a\)\(=2\left(1006+15a\right)\)

Do \(a\) là số nguyên nên ta được: \(2\left(1006+15a\right)⋮2\)

Vậy \(f\left(7\right)-f\left(2\right)\) là hợp số (Đpcm)

f (5)-f(4)=(125a+25b+5c+d)-(64a+19b+4c+d) =61a+9b+c=2012

f(7)-f(2)=(343a+49b+7c+d)-(8a+4b+2c+d)=335a+45b+5c=5(61a+9b+c)+30

=5*(2012+6) chia hết cho 5 mà 5*(2012+6)>5 nên là hợp sô

\(f\left(5\right)-f\left(4\right)=\left(125a+25b+5c+d\right)-\left(64a+16b+4c+d\right)=61a+9b+c=2019\)

\(f\left(7\right)-f\left(2\right)=\left(343a+49b+7c+d\right)-\left(8a+4b+2c+d\right)=335a+45b+5c=5.\left(61a+9b+c\right)+30a=2019+30a⋮3\)

\(\Rightarrowđpcm\)

Lời giải:

Ta có:

\(f(5)-f(4)=2012\)

\(\Leftrightarrow (a.5^3+b.5^2+c.5+d)-(a.4^3+b.4^2+c.4+d)=2012\)

\(\Leftrightarrow 61a+9b+c=2012\)

Do đó:

\(f(7)-f(2)=(a.7^3+b.7^2+c.7+d)-(a.2^3+b.2^2+c.2+d)\)

\(=335a+45b+5c=30a+5(61a+9b+c)\)

\(=30a+5.2012=5(6a+2012)\vdots 5\)

Mà \(f(7)-f(2)=30a+5.2012>5, \forall a\in\mathbb{Z}^+\). Do đó $f(7)-f(2)$ là hợp số (đpcm)

A lộn ngược là mọi ...

Tham khảo:

f(5)=125a+25b+5c+d

f(4)=64a+16b+4c+d

=>f(5)-f(4)=(125a+25b+5c+d)-(64a+16b+4c+d)

=125a+25b+5c+d-64a-16b-4c-d

=61a+9b+c=2019

f(7)=343a+49b+7c+d

f(2)=8a+4b+2c+d

f(7)-f(2)=(343a+49b+7c+d)-(8a+4b+2c+d)

=343a+49b+7c+d-8a-4b-2c-d

=335a+45b+5c

=5(67a+9b+c)

=5(6a+1019) chia hết cho 5

Vậy f(7)-f(2) là hợp số (đpcm)

Ta có : \(f\left(5\right)-f\left(4\right)=2019\Leftrightarrow\left(125a+25b+5c+d\right)-\left(64a+16b+4c+d\right)=2019\)

\(\Leftrightarrow61a+9b+c=2019\left(1\right)\)

Lại có : \(f\left(7\right)-f\left(2\right)=\left(345a+49b+7c+d\right)-\left(8a+4b+2c+d\right)\)

\(=335a+45b+5c=305a+45b+5c+30a=5\left(61a+9b+c\right)+30a\)

\(=2012+30a=2\left(1006+15a\right)⋮2\left(2\right)\)

\(\Rightarrowđpcm\)

Đặt \(g(x)=10x\).

Ta có \(g\left(1\right)=10=f\left(1\right);g\left(2\right)=20=f\left(2\right);g\left(3\right)=30=f\left(3\right)\).

Từ đó \(\left\{{}\begin{matrix}f\left(1\right)-g\left(1\right)=0\\f\left(2\right)-g\left(2\right)=0\\f\left(3\right)-g\left(3\right)=0\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=Q\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\).

\(\Rightarrow f\left(x\right)=10x+Q\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow f\left(8\right)+f\left(-4\right)=80+Q\left(x\right).7.6.5+\left(-40\right)+Q\left(x\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-50=40\).