Question

Cho a + b + c = 0 với a,b,c khác 0. Rút gọn biểu thức:

\(\frac{ab}{a^2+b^2-c^2}\)  +  \(\frac{bc}{b^2+c^2-a^2}\)+   \(\frac{ac}{c^2+a^2-b^2}\)

Answer 1

Ta có: a + b = c <=> a² + b² + 2ab = c² <=> a² + b² - c² = - 2ab

Tương tự: a² + c² - b² = - 2ac

b² + c² - a² = - 2bc

Thế vào ta được

\(\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{a^2+c^2-b^2}=-\frac{ab}{2ab}-\frac{bc}{2bc}-\frac{ac}{2ac}=-6\)

Answer 2

=-6 ngo như bù

Answer 3

a+b+c=0 mà sao a+b=c phải là a+b=-c chứ

Answer 4

Cách khác :

a + b + c = 0

=> b + c = -a

Tương tự có c + a = -b; a + b = -c

\(\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{c^2+a^2-b^2}\)

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

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

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

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

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

\(=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}\)

\(=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}\)

\(=\frac{-3}{2}\)