Question

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

Biết a+b+c=0 . Tính T

Answer 1

ta có: \(T=\frac{a^2}{\left(a-b\right).\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right).\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right).\left(c+a\right)-b^2}\)

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

mà a + b + c = 0 => b + c = -a => b² + 2bc + c² = a²  => a² - b² - c² = 2bc

tương tự như trên, ta có: b²  - c² - a² = 2ac; c² - a² - b² = 2ab

\(\Rightarrow T=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)

Lại có: a+b+c = 0 => a³ + b³ + c³ = 3abc

\(\Rightarrow T=\frac{3abc}{2abc}=\frac{3}{2}\)