Question

nr nat a si b sunt consecutive cu b mai mic decat a
10×a-5×(a-b)×(a+b)
ab-ba

Answer 1

b si a numere consecutive, b<a => a=b+1 si a-b=1

10•a-5•(a-b)•(a+b)

=10a-5•1•(b+1+b)

=10a-5•(2b+1)

=10a-10b-5

=10(a-b) -5

=10•1 -5

=5

ab-ba

=10a+b-10b-a

=9a-9b

=9(a-b)

=9