Bài 1:tính giá trị nhỏ nhất x^2+20y^2+8xy-4y+2021

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!!!