Rất khó

Y

young_wolf

Rất dễ

Cho $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=A$

Ta được:

$A.\dfrac{1}{b-c}=\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)(b-c)}+\dfrac{c}{(b-c)(a-b)}=
\dfrac{a}{(b-c)^2}+\dfrac{ab-b^2}{(a-b)(b-c)(c-a)}+\dfrac{c^2-ca}{(a-b)(b-c)(c-a)}=0$

$A.\dfrac{1}{c-a}=\dfrac{a}{(b-c)(c-a)}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(c-a)(a-b)}=
\dfrac{b}{(c-a)^2}+\dfrac{a^2-ab}{(a-b)(b-c)(c-a)}+\dfrac{bc-c^2}{(a-b)(b-c)
(c-a)}=0$

$A.\dfrac{1}{a-b}=\dfrac{a}{(b-c)(a-b)}+\dfrac{b}{(c-a)(a-b)}+\dfrac{c}{(a-b)^2}=
\dfrac{c}{(a-b)^2}+\dfrac{ac-a^2}{(a-b)(b-c)(c-a)}+\dfrac{b^2-bc}{(a-b)(b-c)
(c-a)}=0$

Cộng 3 vế:

$(\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2})+\dfrac{ab-b^2+c^2-
ca+a^2-ab+bc-c^2+ac-a^2+b^2-bc}{(a-b)(b-c)(c-a)}=0$

$\leftrightarrow \dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}+0=0$

$\leftrightarrow \dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}
{(a-b)^2}=0(dpcm)$

P/S:Đánh toát cả mồ hôi
 
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