Question

Упрастите выражение  a^2/(a-b)(a-c)+(b^2)/(b-c)(b-a)+(c^2)/(c-a)(c-b)
На картинке пример в

Answer 1

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