Question

[Phùng Yến Nga]

Cho a+b+c=0 và abc khác 0. Rút gọn: T = a²/(a-b)(a+b)-c ² + b²/(b-c)(b+c)-a ² + c²/(c-a)(c+a)-b ²

** giúp mk với. mk đang cần gấp

Answer 1

[Tiểu Thiên Yết]

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

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

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

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

\(=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)

\(=\frac{a^3+b^3+c^3}{2abc}\)

Từ \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\) ( tự chứng minh nhé )

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

Vậy T=3/2