Question

Cho 3 số a, b, c khác 0 thõa mãn điều kiện (a + b + c)² = a² + b² + c². Chứng minh rằng:\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}=1\)

Answer 1

(a+b+c)²=a²+b²+c²

=>2(ab+bc+ac)=0

=>ab+bc+ac=0

=> bc=-ab-ac

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

Tuong tu => \(\frac{b^2}{b^2+2ac}=....\)

                     \(\frac{c^2}{c^2+2ab}=...\)

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

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

                                        =1