Question

Cho 3 số tự nhiên a,b,c khác 0 và thoả mãn a+b+c=0. Tính giá trị biểu thức:

\(P=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)

GIÚP MK VS!!!!

Answer 1

\(P=\frac{1}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{1}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{1}{c^2+\left(a+b\right)\left(a-b\right)}\)

\(=\frac{1}{a^2-a\left(b-c\right)}+\frac{1}{b^2-b\left(c-a\right)}+\frac{1}{c^2-c\left(a-b\right)}\)

\(=\frac{1}{a\left(a+c-b\right)}+\frac{1}{b\left(b+a-c\right)}+\frac{1}{c\left(c+b-a\right)}\)

\(=\frac{1}{a\left(-2b\right)}+\frac{1}{b\left(-2c\right)}+\frac{1}{c\left(-2a\right)}=\frac{-\left(a+b+c\right)}{2abc}=0\)

Answer 2

Ta có : a + b + c = 0

=> a + b = - c => a² + 2ab + b² = c²

=> a² + b² - c² = - 2ab

Tương tự : b² + c² - a² = - 2bc

c² + a² - b² = - 2ac

=> \(P=\frac{-1}{2ab}-\frac{1}{2bc}-\frac{1}{2ac}=\frac{-c-a-b}{2abc}=\frac{-\left(a+b+c\right)}{2abc}=0\)