\(\sqrt{x-20}\) + \(\sqrt{y-11}\) + \(\sqrt{z-2020}\)= \(\frac{1}{2}\left(x+y+z\right)\) -1024, Đk: x≥ 20, y≥ 11, z≥ 2020.
⇔ 2\(\sqrt{x-20}\) + 2\(\sqrt{y-11}\) + 2\(\sqrt{z-2020}\) = x+y+z - 2048
⇔x-20 - 2\(\sqrt{x-20}\) +1 + y-11 -2\(\sqrt{y-11}\) +1 +z-2020 - 2\(\sqrt{z-2020}\) +1=0.
⇔ \(\left(\sqrt{x-20}-1\right)^2\) + \(\left(\sqrt{y-11}-1\right)^2\) +\(\left(\sqrt{z-2020}-1\right)^2\) =0
Có: \(\left(\sqrt{x-20}-1\right)^2\)≥ 0 , ∀ x; \(\left(\sqrt{y-11}-1\right)^2\)≥0, ∀ y;
\(\left(\sqrt{z-2020}-1\right)^2\) ≥ 0, ∀ z.
⇒ \(\left(\sqrt{x-20}-1\right)^2\) + \(\left(\sqrt{y-11}-1\right)^2\) +\(\left(\sqrt{z-2020}-1\right)^2\)≥ 0,∀ x,y,z
⇒\(\left\{{}\begin{matrix}\left(\sqrt{x-20}-1\right)^2=0\\\left(\sqrt{y-11}-1\right)^2=0\\\left(\sqrt{z-2020}-1\right)^2=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}\sqrt{x-20}-1=0\\\sqrt{y-11}-1=0\\\sqrt{z-2020}-1=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\sqrt{x-20}=1\\\sqrt{y-11}=1\\\sqrt{z-2020}=1\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=21\left(TMĐK\right)\\y=12\left(TMĐK\right)\\z=2021\left(TMĐK\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm (x;y;z)=(21;12;2021).
