Question

[Dĩnh Bảo]

Cho:

x+y+z =1

x² + y² + z² =1

x³ + y³ + z³ =1

Thính M= x⁸ + y¹¹ + z²⁰¹⁸

Answer 1

[Nguyễn Việt Lâm]

\(x+y=1-z\Rightarrow x^2+y^2+2xy=z^2-2z+1\)

\(\Leftrightarrow1-z^2+2xy=z^2-2z+1\)

\(\Leftrightarrow xy=z^2-z\)

Ta có:

\(1=x^3+y^3+z^3=x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)\)

\(=\left(x+y\right)^3+z^3-3\left(z^2-z\right)\left(1-z\right)\)

\(=\left(1-z\right)^3+z^3-3\left(z^2-z\right)\left(1-z\right)\)

\(=1-3z+3z^2-z^3+z^3-3\left(2z^2-z^3-z\right)\)

\(=3z^3-3z^2+1\)

\(\Rightarrow3z^3-3z^2=0\)\(\Rightarrow\left[{}\begin{matrix}z=0\\z=1\end{matrix}\right.\)

- Với \(z=0\Rightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)

- Với \(z=1\Rightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị

\(\Rightarrow M=1\)

Answer 2

[Dĩnh Bảo]

.mình k hiểu