Giải thích các bước giải:
$A=x^2+5x+1$
$=x^2+\dfrac{5}{2}.2.x+(\dfrac{5}{2})^2-(\dfrac{5}{2})^2+1$
$=(x+\dfrac{5}{2})^2-\dfrac{21}{4}$
$\text{Ta có:}$
$(x+\dfrac{5}{2})^2≥0$ $∀x∈R$
$⇒(x+\dfrac{5}{2})^2-\dfrac{21}{4}≥-\dfrac{21}{4}$ $∀x∈R$
$\text{Dấu "=" xảy ra khi:}$
$(x+\dfrac{5}{2})^2-\dfrac{21}{4}=-\dfrac{21}{4}$
$⇔(x+\dfrac{5}{2})^2=0$
$⇔x+\dfrac{5}{2}=0$
$⇔x=-\dfrac{5}{2}$
$\text{Vậy $Min_{(A)}=-\dfrac{21}{4}$ tại $x=-\dfrac{5}{2}$}$
$B=x^2+2y^2-6y+2x-2xy+2020$
$⇔x^2-2xy+y^2+y^2-4y+4+2x-2y+2016$
$⇔(x-y)^2+(y-2)^2+2(x-y)+2016$
$⇔[(x-y)^2+2(x-y)+1]+(y-2)^2+2015$
$⇔(x-y+1)^2+(y-2)^2+2015$
$\text{Ta có:}$
$(x-y+1)^2≥0$ $∀x;y∈R$
$(y-2)^2≥0$ $∀y∈R$
$⇒(x-y+1)^2+(y-2)^2+2015≥2015$ $∀x;y∈R$
$\text{Dấu "=" xảy ra khi:}$
$(x-y+1)^2+(y-2)^2+2015=2015$
$⇔(x-y+1)^2+(y-2)^2=0$
$⇔$ \(\left[ \begin{array}{l}x-y+1=0\\y-2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x-y=-1\\y=2\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x-2=-1\\y =2\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=1\\y=2\end{array} \right.\)
$\text{Vậy $Min_{(A)}=2015$ tại $x=1;y=2$}$
Học tốt!!!
