\(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)
\(\Rightarrow\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)}\)
\(\Leftrightarrow\dfrac{a^2}{\left(b-c\right)^2}=\dfrac{ab^2-a^2b+a^2c-ac^2}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)
Tương tự ta có:
\(\dfrac{b^2}{\left(c-a\right)^2}=\dfrac{bc^2-b^2c+b^2a-a^2b}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\)
\(\dfrac{c^2}{\left(a-b\right)^2}=\dfrac{a^2c-c^2a+c^2b-cb^2}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Cộng 3 đẳng thức trên có:
==" xl mk ko bt tài làm để có bình phương đc :)) mk chỉ can chứng minh
\(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0đcthui\)
