Question

Cho a+b+c=1. Tính  E = a^3/((a-b)*(a-c)) + b^3/((b-c)*(b-a)) + c^3/((c-a)*(c-b))

Answer 1

Ta có \(E=\frac{a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=a+b+c=1\)

Ở đây chú ý rằng \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=-\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)\)