Cho \(x,y,z \in \mathbb R^+\) thỏa mãn: \(\left\{\begin{matrix}
x+xy+y=1 & \\
y+yz+z=3& \\
z+xz+x=7&
\end{matrix}\right.\)
Tính Tính \(M = z+y^{2010}+x^{2020}\)
Cho \(x,y,z \in \mathbb R^+\) thỏa mãn: \(\left\{\begin{matrix}
x+xy+y=1 & \\
y+yz+z=3& \\
z+xz+x=7&
\end{matrix}\right.\)
Tính Tính \(M = z+y^{2010}+x^{2020}\)
Ta có : : $\left\{ \begin{array}{l}x+xy+y=1\\y+yz+z=3\\z+zx+z=7\end{array} \right.$
$\to \left\{ \begin{array}{l}(x+1).(y+1) = 2\\(y+1).(z+1) = 4\\(z+1).(x+1) = 8\end{array} \right.$ $(*)$
$\to [(x+1).(y+1).(z+1)]^2 = 64$
$\to \left[ \begin{array}{l}(x+1).(y+1).(z+1) = 8\\(x+1).(y+1).(z+1) = -8\end{array} \right.$
Mà : $x,y,z \in \mathbb{R^+}$ do đó $(x+1).(y+1).(z+1) = 8$
Kết hợp từ $(*)$ suy ra :
$\left\{ \begin{array}{l}z+1=4\\x+1=2\\y+1=1\end{array} \right.$
$\to z=3,x=1,y=0$
Khi đó ta có : $M = z+y^{2010} + x^{2020}$
$ = 3+ 0^{2010} + 1^{2020} = 3+0+1 =4$
Vậy $M=4$