a: ĐKXĐ: \(\left(a+b+c\right)^2-\left(ab+bc+ca\right)<>0\)
=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)-\left(ab+ac+bc\right)<>0\)
=>\(a^2+b^2+c^2+ab+ac+bc<>0\)
=>\(2a^2+2b^2+2c^2+2\left(ab+ac+bc\right)<>0\)
=>\(\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)<>0\)
=>\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2<>0\)
Dấu '=' xảy ra khi \(\begin{cases}a+b=0\\ b+c=0\\ a+c=0\end{cases}\Rightarrow a=b=c=0\)
=>Để M xác định thì \(a^2+b^2+c^2<>0\)
b: \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+ac+bc\right)^2\)
\(=\left(a^2+b^2+c^2\right)\left\lbrack a^2+b^2+c^2+2\left(ab+ac+bc\right)\right\rbrack+\left(ab+ac+bc\right)^2\)
\(=\left(a^2+b^2+c^2\right)^2+2\left(a^2+b^2+c^2\right)\left(ab+ac+bc\right)+\left(ab+ac+bc\right)^2\)
\(=\left(a^2+b^2+c^2+ab+ac+bc\right)^2\)
\(\left(a+b+c\right)^2-ab-ac-bc\)
\(=a^2+b^2+c^2+2\left(ab+ac+bc\right)-\left(ab+ac+bc\right)\)
\(=a^2+b^2+c^2+ab+ac+bc\)
Ta có: \(M=\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+ac+bc\right)^2}{\left(a+b+c\right)^2-ab-ac-bc}\)
\(=\frac{\left(a^2+b^2+c^2+ab+ac+bc\right)^2}{\left(a^2+b^2+c^2+ab+ac+bc\right)}\)
\(=a^2+b^2+c^2+ab+ac+bc\)
