Cho a, b, c đôi một khác nhau và khác 0 không thỏa mãn: (a+b+c)^2 = a^2 + b^2 + c^2 Tính giá trị biểu thức: A = a^2/(a^2+2bc) + b^2/(b^2+2ca) + c^2/(c^2+2ab)

$(a+b+c)^2=a^2+b^2+c^2$

$=>a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2$

$=>ab+bc+ca=0$

$=>ab=-bc-ca, bc=-ab-ac, ca=-ab-bc$

$a^2+2bc$

$=a^2+bc-ab-ac$

$=a(a-b)-c(a-b)$

$=(a-b)(a-c)$

Tương tự: $b^2+2ca=(b-c)(b-a)=-(a-b)(b-c)$

$c^2+2ab=(c-a)(c-b)=(a-c)(b-c)$

$=>A=\dfrac{a^2}{(a-b)(a-c)}-\dfrac{b^2}{(a-b)(b-c)}+\dfrac{c^2}{(a-c)(b-c)}$

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

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

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

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

$=\dfrac{(b-c)(a-c)(a-b)}{(a-b)(a-c)(b-c)}$

$=1$