Question

Cho 3 số thực a, b, c đôi một khác nhau thỏa mãn: \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)

CMR: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)

Answer 1

Đặt x = a - b, y = b - c, z = c - a

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\ay+bz+cx=ab-ac+bc-ab+ac-bc=0\end{matrix}\right.\)

+ \(ay+bz+cx=0\)

\(\Rightarrow\dfrac{1}{y}\left(\dfrac{a}{y}+\dfrac{b}{z}+\dfrac{c}{x}\right)=0\)

\(\Rightarrow\dfrac{a}{y^2}+\dfrac{bx}{xyz}+\dfrac{cz}{xyz}=0\)

\(\Rightarrow\dfrac{a}{y^2}=\dfrac{-bx-cz}{xyz}\)

+ Tương tự : \(\dfrac{b}{z^2}=\dfrac{-cy-ax}{xyz}\)

\(\dfrac{c}{x^2}=\dfrac{-az-by}{xyz}\)

Do đó : \(\dfrac{a}{y^2}+\dfrac{b}{z^2}+\dfrac{c}{x^2}=\dfrac{-a\left(x+z\right)-b\left(x+y\right)-c\left(y+z\right)}{xyz}\)

\(=\dfrac{ay+bz+cx}{xyz}\) ( do x + y + z = 0)

\(=0\) ( do ay + bz + cx = 0 )