Question

Cho f(x)=ax^3+bx^2+cx+d

Biết f(5)-f(4)=2019

a là số nguyên dương

CMR f(7)-f(2) là hợp số

Answer 1

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)

Answer 2

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

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

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

\(\Rightarrowđpcm\)