Đáp án:
Giải thích các bước giải:
$y=2020+\sqrt{2x^2-4x+3}$
$=\sqrt{2\left(x^2-2x+\dfrac{3}{2}\right)}+2020$
$=\sqrt{2\left(x^2-2x+1+\dfrac{1}{2}\right)}+2020$
$=\sqrt{2(x-1)^2+1}+2020$
Ta có:
$2(x-1)^2\ge 0$
$⇒2(x-1)^2+1\ge 1$
$⇒\sqrt{2(x-1)^2+1}\ge 1$
$⇒\sqrt{2(x-1)^2+1}+2020\ge 2021$
$⇒y\ge 2021⇒y_{\min}=2021$
Dấu "=" xảy ra khi:
$2(x-1)^2=0$
$⇒x-1=0$
$⇒x=1$
Vậy $y_{\min}=2021$ khi $x=1$.
