Question

\(\sqrt{x+2018}+\sqrt{y-2019}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\)

Answer 1

Lời giải:

PT \(\Leftrightarrow 2\sqrt{x+2018}+2\sqrt{y-2019}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow (x+2018-2\sqrt{x+2018}+1)+(y-2019-2\sqrt{y-2019}+1)+(z-2-2\sqrt{z-2}+1)=0\)

\(\Leftrightarrow (\sqrt{x+2018}-1)^2+(\sqrt{y-2019}-1)^2+(\sqrt{z-2}-1)^2=0\)

Vì \((\sqrt{x+2018}-1)^2\geq 0; (\sqrt{y-2019}-1)^2\geq 0; (\sqrt{z-2}-1)^2\geq 0\). Do đó để tổng của chúng bằng $0$ thì:

\((\sqrt{x+2018}-1)^2=(\sqrt{y-2019}-1)^2=(\sqrt{z-2}-1)^2= 0\)

\(\Rightarrow \left\{\begin{matrix} x=-2017\\ y=2020\\ z=3\end{matrix}\right.\)