Question

Cho a+b+c=0 và a,b,c\(\ne0\) . Chứng minh rằng:

A=\(\sqrt{\dfrac{6a^2}{a^2-b^2-c^2}+\dfrac{6b^2}{b^2-c^2-a^2}+\dfrac{6c^2}{c^2-a^2-b^2}}\) là số nguyên

Answer 1

Ta có:

\(a^2=\left(-b-c\right)^2\)

\(\Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự ta cũng có

\(\left\{{}\begin{matrix}b^2-c^2-a^2=2ca\\c^2-a^2-b^2=2ab\end{matrix}\right.\)

Thế vô ta được

\(A=\sqrt{\dfrac{3a^2}{bc}+\dfrac{3b^2}{ca}+\dfrac{3c^2}{ab}}\)

\(=\sqrt{\dfrac{3\left(a^3+b^3+c^3\right)}{abc}}\)

\(=\sqrt{3.\dfrac{\left(a^3+b^3+c^3-3abc\right)+3abC}{abc}}\)

\(=\sqrt{3.\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{abc}}\)

\(=\sqrt{3.3}=3\)

ĐPCM