Đáp án: $P=1$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac1c=0$
$\to bc+ca+ab=0$
$\to bc=-ab-ac$
$\to 2bc=bc-ab-ac$
$\to a^2+2bc=a^2+bc-ab-ac$
$\to a^2+2bc=(a^2-ab)+(bc-ac)$
$\to a^2+2bc=a(a-b)-c(a-b)$
$\to a^2+2bc=(a-b)(a-c)$
$\to\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{(a-b)(a-c)}$
Tương tự
$\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{(b-c)(b-a)}$
$\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{(c-a)(c-b)}$
$\to P=\dfrac{a^2}{(a-b)(a-c)}+\dfrac{b^2}{(b-c)(b-a)}+\dfrac{c^2}{(c-a)(c-b)}$
$\to -P=\dfrac{a^2}{(a-b)(c-a)}+\dfrac{b^2}{(a-b)(b-c)}+\dfrac{c^2}{(b-c)(c-a)}$
$\to -P=\dfrac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{a^2(b-c)+b^2c-b^2a+c^2a-c^2b}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{a^2(b-c)+(b^2c-c^2b)-(b^2a-c^2a)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{a^2(b-c)+bc(b-c)-a(b^2-c^2)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{a^2(b-c)+bc(b-c)-a(b-c)(b+c)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{(b-c)(a^2+bc-a(b+c))}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{(b-c)(a^2+bc-ab-ca)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{(b-c)((a^2-ab)+(bc-ca)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{(b-c)(a(a-b)-c(a-b)}{(a-b)(b-c)(c-a)}$
$\to -P=\dfrac{(b-c)(a-c)(a-b)}{(a-b)(b-c)(c-a)}$
$\to -P=-1$
$\to P=1$
