Sửa đề: \(M=2019\sqrt{x-2}+2020\sqrt{10-y}\)
+Có: \(\sqrt{x-2}\ge với\forall x\\ \sqrt{10-y}\ge0với\forall x\\ \Rightarrow2019\sqrt{x-2}+2020\sqrt{10-y}\ge0\\ \Leftrightarrow M\ge0\)
+Dấu ''='' xảy ra khi
\(\sqrt{x-2}=0\\ \Leftrightarrow x=2\)
\(\sqrt{10-y}=0\\ \Leftrightarrow y=10\)
+Vậy \(M_{min}=0\) khi \(x=2,y=10\)