Question

Cho a, b, c thỏa mãn:

\(\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-a\right)\left(b-c\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}=2013\)

Tính giá trị của biểu thức:

\(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\)

Answer 1

Lời giải:
Ta có:

\(\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-a)(b-c)}+\frac{a-b}{(c-a)(c-b)}=2013\)

\(\Leftrightarrow \frac{-(b-c)^2}{(a-b)(b-c)(c-a)}+\frac{-(c-a)^2}{(a-b)(b-c)(c-a)}+\frac{-(a-b)^2}{(a-b)(b-c)(c-a)}=2013\)

\(\Leftrightarrow \frac{-[(a-b)^2+(b-c)^2+(c-a)^2]}{(a-b)(b-c)(c-a)}=2013\)

\(\Rightarrow \frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}=-2013(*)\)

Lại có:

\(P=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)

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

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

\(=\frac{ab+bc+ac-(a^2+b^2+c^2)}{(a-b)(b-c)(c-a)}=-\frac{1}{2}.\frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}\)

\(=\frac{-1}{2}.-2013=\frac{2013}{2}\) (theo $(*)$)