Question

cho x,y,z thỏa mãn \(x^2\)+2\(y^2\)+\(z^2\)-2xy-2y-4z+5=0

tính A=\(\left(x-1^{ }\right)^{2015}\)+\(\left(y-1\right)^{2015}\)+\(\left(z-1\right)^{2015}\)

Answer 1

Lời giải:

Ta có:

\(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

\(\Leftrightarrow (x^2+y^2-2xy)+(y^2-2y+1)+(z^2-4z+4)=0\)

\(\Leftrightarrow (x-y)^2+(y-1)^2+(z-2)^2=0\)

Ta thấy:

\(\left\{\begin{matrix} (x-y)^2\geq 0\\ (y-1)^2\geq 0\\ (z-2)^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)

\(\Rightarrow (x-y)^2+(y-1)^2+(z-2)^2\geq 0\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-y=0\\ y-1=0\\ z-2=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=1\\ y=1\\ z=2\end{matrix}\right.\)

Do đó:

\(A=(x-1)^{2015}+(y-1)^{2015}+(z-1)^{2015}=1\)