\(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
Viết lại đề như sau: \(\hept{\begin{cases}x+y+z=3\\2xy-z^2=9\end{cases}}\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz-2xy+z^2=0\)
\(\Leftrightarrow x^2+y^2+2z^2+2yz+2xz=0\)
\(\Leftrightarrow\left(x+z\right)^2+\left(y+z\right)^2=0\)
\(\Leftrightarrow x=y=-z\Leftrightarrow\frac{1}{a}=\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow a=b=-c\)
\(M=\left(a-3b+c\right)^{2018}=\left(a-3a-a\right)^{2018}=\left(3a\right)^{2018}\)