\(vt:\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2+10^2}\)

=\(\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+..+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)

=>A<1

ta có f(x)=ax\(^2\)+bx+c

tại x=0 =>f(0)=c\(⋮\)7(1)

x=1=>f(1)=a+b+c\(⋮\)7

mà c\(⋮\)7=>a+b\(⋮\)7(2)

x=-1=>f(-1)=a-b+c

mà c\(⋮\)7=>a-b\(⋮\)7(3)

từ (2)(3)có a+b+a-b=2a\(⋮\)7

mà 2;7=(1)

=>a\(⋮\)7(4)

từ (4)(3)ta có a-b\(⋮\)7

a\(⋮\)7

=>b\(⋮\)7(5)

từ (1)(4)(5)suy ra a,b,c\(⋮\)7

b