\(3=a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\Rightarrow a+b+c\le3\)

\(M=2\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(a+b+c\right)+\dfrac{9}{a+b+c}\)

\(=2\left[a+b+c+\dfrac{9}{a+b+c}\right]-\dfrac{9}{a+b+c}\ge2.\sqrt{9}-\dfrac{9}{3}=6-3=3\)Min = 3 khi a=b=c =1

\(\left\{{}\begin{matrix}x+y=-1\\xy=-\dfrac{1}{4}\end{matrix}\right.\)

\(x^2+y^2=\left(x+y\right)^2-2xy=1+\dfrac{1}{2}=\dfrac{3}{2}\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1-3.\dfrac{1}{4}.1=\dfrac{1}{4}\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=\dfrac{9}{4}-2.\dfrac{1}{16}=\dfrac{17}{8}\)

\(x^7+y^7=\left(x^4+y^4\right)\left(x^3+y^3\right)-x^3y^3\left(x+y\right)=\dfrac{17}{8}.\dfrac{1}{4}-\dfrac{1}{4^3}.\left(-\dfrac{1}{4}\right)=\dfrac{1}{4}.\dfrac{35}{64}=\dfrac{35}{256}.\)

x=1