Question

[Duong Thi Nhuong]

Cho x, y, z thõa mãn đồng thời: \(3x-2y-2\sqrt{y+2012}+1=0;3y-2z-2\sqrt{z-2013}+1=0;3z-2x-z\sqrt{x-2}-2=0\)

Answer 1

[soyeon_Tiểubàng giải]

Cộng vế với vế của 3 đẳng thức đã cho ta được:

\(x+y+z-2\sqrt{y+2012}-2\sqrt{z-2013}-2\sqrt{x-2}=0\)

\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2012-2\sqrt{y+2012}+1\right)+\left(z-2013+2\sqrt{z-2013}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-1\right)^2=0\\\left(\sqrt{y+2012}-1\right)^2=0\\\left(\sqrt{z-2013}-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y+2012}-1=0\\\sqrt{z-2013}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y+2012}=1\\\sqrt{z-2013}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2011\\z=2014\end{matrix}\right.\)

Thay vào C ta được:

C = (3 - 4)²⁰¹⁶ + (-2011 + 2012)²⁰¹⁷ + (2014 - 2013)²⁰¹⁸

C = 1 + 1 + 1 = 3

Answer 2

[Duong Thi Nhuong]

THÊM

Cho x, y, z thõa mãn đồng thời:

\(3x-2y-2\sqrt{y+2012}+1=0;3y-2z-2\sqrt{z-2013}+1=0;3z-2x-2\sqrt{x-2-2=0.}\)Tính \(C=\left(x-4\right)^{2016}+\left(y+2012\right)^{2017}+\left(z-2013\right)^{2018}\)