ta có x2+5x+4

=x2+x+4x+4

=(x2+x)+(4x+4)

=x(x+1)+4(x+1)

=(x+1)(x+4)

tương tự ta đc

x2+11x+28=(x+4)(x+7)

x2+17x+70=(x+7)(x+10)

x2+23x+130=(x+10)(x+13)

=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+-+\dfrac{1}{x+13}=\dfrac{4}{13}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)

=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)

=> 13(x+13)-13(x+1)=4(x+1)(x+13)

=> 13[(x+13)-(x+1)]=(4x+4)(x+13)

=>13(x+13-x-1)=4x2+52x+4x+52

=13.12=4x2+56x+52

=>4x2+56x+52=156

=>4x2+56x-104=0