GTNN của biểu thức : A= (x-1)^2021 + (x-2)^2022

Là   MAX A = 1  khi  \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

cho 10 k

\(D=\left(x+1\right)\left(x+4\right)\left(x^2+5x+8\right)+2021\)

    \(=\left(x^2+5x+4\right)\left(x^2+5x+8\right)+2021\)

Đặt \(x^2+5x+6=t\)

Ta có: \(D=\left(t-2\right)\left(t+2\right)+2021\)

              \(=t^2-4+2021=t^2+2017\ge2017\forall t\)

Dấu "=" xảy ra khi: \(t=0\)

\(\Rightarrow x^2+5x+6=0\)

\(\Rightarrow\left(x+2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)

Vậy GTNN cua D là 2017 khi \(\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)

Chúc bạn học tốt.

?

\(P=x^2+4y^2-4x+4y+2021\)

\(=\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)+2016\)

\(=\left(x-2\right)^2+\left(2y+1\right)^2+2016\ge2016\)

\(P_{min}=2016\Leftrightarrow x=2;y=-\dfrac{1}{2}\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

rất tiếc em mới học lớp 6

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Lời giải:
$A=x^2+2x+2xy+2y^2+4y+2021$

$=(x^2+2xy+y^2)+2x+y^2+4y+2021$

$=(x+y)^2+2(x+y)+1+(y^2+2y+1)+2019$

$=(x+y+1)^2+(y+1)^2+2019\geq 2019$

Vậy $A_{\min}=2019$ khi $x+y+1=y+1=0$

$\Leftrightarrow (x,y)=(0,-1)$