Lời giải:
Đặt $\sqrt{y}=b(b\geq 0)\Rightarrow y=b^2$
$M=2x^2+5b^2-4xb-4x-8b+2036$
$=2(x^2+b^2-2xb)+3b^2-4x-8b+2036$
$=2(x-b)^2-4(x-b)+3b^2-12b+2036$
$=2(x-b)^2-4(x-b)+2+3(b^2-4b+4)+2022$
$=2[(x-b)^2-2(x-b)+1]+3(b-2)^2+2022$
$=2(x-b-1)^2+3(b-2)^2+2022\geq 2022$
Vậy $M_{\min}=2022$