Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ac=0\)
Khi đó:
\(A=\frac{a^2}{a^2+bc-ab-ac}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ac-bc}\)
\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)}\)
\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{a^2(c-b)-b^2[(c-b)+(b-a)]+c^2(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(a^2-b^2)(c-b)+(c^2-b^2)(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(a-b)(c-b)[(a+b)-(c+b)]}{(a-b)(b-c)(c-a)}=\frac{(a-b)(c-b)(a-c)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=1\)
Đúng 0
Bình luận (0)
