Giải thích các bước giải:
$x^2+20y^2+8xy-4y+2021$
$=x^2+8xy+16y^2+4y^2-4y+1+2020$
$=(x+4y)^2+(2y-1)^2+2020$
$\text{Ta có:}$
$(x+4y)^2≥0$ $∀x;y∈R$
$(2y-1)^2≥0$ $∀y∈R$
$⇒(x+4y)^2+(2y-1)^2+2020≥2020$ $∀x;y∈R$
$\text{Dấu "=" xảy ra khi}$
$(x+4y)^2+(2y-1)^2+2020=2020$
$⇔(x+4y)^2+(2y-1)^2=0$
$⇔$ \(\left[ \begin{array}{l}(2y-1)^2=0\\(x+4y)^2=0\end{array} \right.\)$⇔$ \(\left[ \begin{array}{l}2y-1=0\\x+4y=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}2y=1\\x+4y=0\end{array} \right.\)$⇔$ \(\left[ \begin{array}{l}y=\dfrac{1}{2}\\x+4y=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}y=\dfrac{1}{2}\\x=0-4.\dfrac{1}{2}\end{array} \right.\)$⇔$ \(\left[ \begin{array}{l}y=\dfrac{1}{2}\\x=-2\end{array} \right.\)
$\text{Vậy $Min_{(A)}=2020$ tại $x=-2;y=\dfrac{1}{2}$}$
Học tốt!!!
