Giải thích các bước giải:
$A=(x-2)^2+|y-5|+3$
$\text{Ta có:}$
$(x-2)^2≥0$ $ ∀x∈Q$
$|y-5|≥ 0$ $∀y∈Q$
$⇒(x-2)^2+|y-5|+3≥3$ $∀x;y∈Q$
$\text{Dấu '=' xảy ra khi:}$
$(x-2)^2+|y-5|+3=3$
$⇔(x-2)^2+|y-5|=0$
$⇔$ \(\left[ \begin{array}{l}(x-2)^2=0\\|y-5|=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x-2=0\\y-5=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=2\\y=5\end{array} \right.\)
$\text{Vậy $Min(A)=3$ tại $x=2;y=5$}$
$B=(y-2020)^2+(x-2021)^4-3$
$\text{Ta có:}$
$(y-2020)^2≥0$ $∀y∈R$
$(x-2021)^4≥0$ $∀x∈R$
$⇒(y-2020)^2+(x-2021)^4-3≥-3$ $∀x;y∈R$
$\text{Dấu '=' xảy ra khi:}$
$(y-2020)^2+(x-2021)^4-3=-3$
$⇔(y-2020)^2+(x-2021)^4=0$
$⇔$ \(\left[ \begin{array}{l}(y-2020)^2=0\\(x-2021)^4=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}y-2020=0\\x-2021=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}y=2020\\x=2021\end{array} \right.\)
$\text{Vậy $Min(B)=-3$ tại $x=2021;y=2020$}$
Học tốt!!!
