Question

Giúp mình với

Answer 1

\(\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}\)

=>\(\left(a+b\right)\left(a+d\right)=\left(b+c\right)\left(c+d\right)\)

=>\(a^2+ad+ab+bd=bc+bd+c^2+cd\)

=>\(a^2-c^2+ad+ab+bd-bc-bd-cd=0\)

=>\(\left(a-c\right)\left(a+c\right)+\left(ad-cd\right)+\left(ab-bc\right)=0\)

=>\(\left(a-c\right)\left(a+c\right)+d\left(a-c\right)+b\left(a-c\right)=0\)

=>\(\left(a-c\right)\left(a+c+b+d\right)=0\)

=>a+b+c+d=0