Toán Lớp 8: Cho `a/(b-c)+b/(c-a)+c/(a-b)=0` `(a≠b≠c)` Tính `P=a/(b-c)^2+b/(c-a)^2+c/(b-a)^2`

Question

Toán Lớp 8:  Cho a/(b-c)+b/(c-a)+c/(a-b)=0 (a≠b≠c) Tính P=a/(b-c)^2+b/(c-a)^2+c/(b-a)^2

haianhtu97 · Answer

$  $    Giải đáp + giải th&iacute;ch c&aacute;c bước giải :    a/(b-c)+b/(c-a)+c/(a-b)=0   -> a/(b-c)=-(b/(c-a)+c/(a-b))   -> a/(b-c)=-((b(a-b))/((a-b)(c-a)) + (c(c-a))/((a-b)(c-a)))   ->a/(b-c)=-( (ab - b^2 + c^2 - ca)/((a-b)(c-a)))   ->a/(b-c) = (-ab+b^2-c^2+ca)/((a-b)(c-a))   ->a/(b-c) = (b^2 - ab+ca-c^2)/((a-b)(c-a))   ->a/(b-c) . 1/(b-c) = (b^2 - ab+ca-c^2)/((a-b)(c-a)) . 1/(b-c)   -> a/(b-c)^2 =(b^2-ab+ca-c^2)/((a-b)(c-a)(b-c))   a/(b-c)+b/(c-a)+c/(a-b)=0   ->b/(c-a)=-(a/(b-c)+c/(a-b))   -> b/(c-a)=-((a(a-b))/((b-c)(a-b))+(c(b-c))/((b-c)(a-b)))   -> b/(c-a) = - (a^2 - ab + bc - c^2)/((b-c)(a-b))   ->b/(c-a) =(c^2 + ab - bc - a^2)/((b-c)(a-b))   ->b/(c-a) . 1/(c-a) = (c^2 + ab - bc - a^2)/((b-c)(a-b)) . 1/(c-a)   ->b/(c-a)^2=(c^2 -bc+ab-a^2)/((a-b)(c-a)(b-c))    a/(b-c)+b/(c-a)+c/(a-b)=0   ->c/(a-b)=-(a/(b-c)+b/(c-a))   ->c/(a-b)=-((a(c-a))/((b-c)(c-a)) + (b(b-c))/((b-c)(c-a)))   ->c/(a-b) = -(ac - a^2 + b^2-bc)/((b-c)(c-a))   ->c/(a-b) . 1/(a-b)=(-ac +a^2 - b^2+bc)/((b-c)(c-a)) . 1/(a-b)   ->c/(a-b)^2=(a^2 - ac - b^2 +bc)/((a-b)(b-c)(c-a))   P=a/(b-c)^2+b/(c-a)^2+c/(b-a)^2   ->P=(b^2-ab+ca-c^2)/((a-b)(c-a)(b-c)) + (c^2 -bc+ab-a^2)/((a-b)(c-a)(b-c)) + (a^2 - ac - b^2 +bc)/((a-b)(b-c)(c-a))   -> P=0/((a-b)(b-c)(c-a))   ->P=0   Vậy P=0