Ta có: 

`(a - b)^2 + (b - c)^2 + (c - a)^2 >= 0` với mọi `a;b;c`

`=> a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 >= 0`

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

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

`=> 3(a^2+b^2+c^2)>=(a+b+c)^2`

`=> đpcm`

\(3x^2-12xy+12y^2\\ =3\left(x^2-4xy+4y^2\right)\\ =3\left(x-2y\right)^2\)

a,DK: \(x\ne\pm4;x\ne-5\)
\(B=\dfrac{2x}{x+4}+\dfrac{16x}{x^2-16}\)
\(B=\dfrac{2x}{x+4}+\dfrac{16x}{\left(x-4\right).\left(x+4\right)}\)
\(B=\dfrac{2x.\left(x-4\right)}{\left(x-4\right).\left(x+4\right)}+\dfrac{16x}{\left(x-4\right).\left(x+4\right)}\)
\(B=\dfrac{2x^2-8x+16x}{\left(x-4\right).\left(x+4\right)}\)
\(B=\dfrac{2x^2+4x}{\left(x-4\right).\left(x+4\right)}\)\(=\dfrac{2x.\left(x+4\right)}{\left(x-4\right).\left(x+4\right)}=\dfrac{2x}{x-4}\)
Vay \(B=\dfrac{2x}{x-4}\)
b, Ta co: \(A.B=\dfrac{x-4}{x+5}.\dfrac{2x}{x-4}=\dfrac{2x}{x+5}\)\(=\dfrac{2x+10-10}{x+5}=\dfrac{2.\left(x+5\right)}{x+5}-\dfrac{10}{x+5}=2-\dfrac{10}{x+5}\)
De A.B la so nguyen \(\Leftrightarrow2-\dfrac{10}{x+5}\in Z\) 
                                  \(\Leftrightarrow\dfrac{10}{x+5}\in Z\)
                                  \(\Leftrightarrow10⋮x+5\)
                                  \(\Leftrightarrow x+5\in\left\{\pm1;\pm2;\pm5\pm10\right\}\)
                                  \(\Leftrightarrow x\in\left\{-6;-3;-7;0;-10;-15\right\}\) do \(x\ne\left\{\pm4;5\right\}\)
Vay x \(\in\left\{-6;-3;-7;0;-10;-15\right\}\)