Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\)
\(\Rightarrow z=-x-y\)
\(M=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=\dfrac{x^3+y^3+z^3}{xyz}\)
\(=\dfrac{x^3+y^3+\left(-x-y\right)^3}{xy\left(-x-y\right)}=\dfrac{x^3+y^3-x^3-y^3-3x^2y-3xy^2}{xy\left(-x-y\right)}\)
\(=\dfrac{-3xy\left(x+y\right)}{-xy\left(x+y\right)}=3\)