Question

Cho \(a,b,c\) đôi một khác nhau và thỏa mãn: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

Tính giá trị biểu thức:

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

Answer 1

Từ \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\)

\(\Rightarrow2ab+2bc+2ca=0\Rightarrow ab+bc+ca=0\)

\(\Rightarrow bc=-ac-ca \Rightarrow a^2+2bc=a^2+bc-ca-ab\)

\(=\left(a-c\right)\left(a-b\right)\). Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(a-c\right)\left(b-c\right)\)

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

\(=\sqrt{\dfrac{a^2}{(a-b)(a-c) }+\dfrac{b^2}{(b-a)(b-c)}+\dfrac{c^2}{(a-c)(b-c)}}=1\)

Answer 2

ngonhuminh

Ace Legona

Hu Hu!