Question

ai giúp em với

Answer 1

Theo đề: \(a+b+c=0\Leftrightarrow c=-a-b\)

\(\dfrac{b-c}{a\left(a-b\right)}+\dfrac{c-a}{b\left(a-b\right)}=\dfrac{b\left(b-c\right)+a\left(c-a\right)}{ab\left(a-b\right)}\\ =\dfrac{b^2-bc+ac-a^2}{ab\left(a-b\right)}=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)}\\ =\dfrac{\left(b-a\right)\left(b+a-c\right)}{ab\left(a-b\right)}=\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab\left(a-b\right)}\\ =\dfrac{c-a-b}{ab}=\dfrac{c+c}{ab}=\dfrac{2c}{ab}\)

 

Answer 2

\(a+b+c=0\Rightarrow c=-a-b\)

\(VT=\dfrac{b-c}{a\left(a-b\right)}+\dfrac{c-a}{b\left(a-b\right)}=\dfrac{b\left(b-c\right)+a\left(c-a\right)}{ab\left(a-b\right)}=\dfrac{b^2-bc+ac-a^2}{ab\left(a-b\right)}=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)}=\dfrac{\left(b-a\right)\left(b+a-c\right)}{ab\left(a-b\right)}=\dfrac{c-a-b}{ab}=\dfrac{c-\left(-c\right)}{ab}=\dfrac{2c}{ab}=VP\)