\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

Bạn coi lại đề, nhìn 2 vế của điều kiên đều là \(\sqrt{x+2}\) có vẻ sai sai rồi đó

Không viết lại đề

\(B=\left|2-x\right|+\left|3-y\right|+\left|x+y-2020\right|\ge\left|2-x+3-y+x+y-2020\right|=\left|-2015\right|=2015\)Còn lại tự làm nốt

Ta có: \(\left(x-3\right)^2\ge0;\left(y-1\right)^2\ge0\)

=> \(B=\left(x-3\right)^2+\left(y-1\right)^2+2020\ge2020\)

Dấu "=" xảy ra <=> x - 3 = 0 và y - 1 = 0 <=> x = 3 và y = 1 

Vậy GTNN của B = 2020 đạt tại x = 3 và y = 1.

\(B=\left(x-3\right)^2+\left(y-1\right)^2+2020\)

Ta có \(\hept{\begin{cases}\left(x-3\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-3\right)^2+\left(y-1\right)^2+2020\ge2020\)

=> B\(\ge\)2020

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-3\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-3=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=1\end{cases}}}\)

Vậy GTNN của B=2020 đạt được khi x=3 và y=1

Đặt \(A=x^2+15y^2+xy+8x+y+2020\)

\(\Rightarrow4A=4x^2+60y^2+4xy+32x+4y+8080\)

\(=\left(4x^2+4xy+y^2\right)+59y^2+32x+4y+8080\)

\(=\left(2x+y\right)^2+16.\left(2x+y\right)+64+59y^2+4y-16y+8016\)

\(=\left(2x+y+8\right)^2+59y^2-12y+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y^2-\frac{59}{12}y\right)+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y^2-2\cdot y\cdot\frac{59}{24}+\frac{59^2}{24^2}-\frac{59^2}{24^2}\right)+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y-\frac{59}{24}\right)^2+7659,439236\ge7659,439236\)

\(\Rightarrow A\ge1914,859809\)

Dấu "=" xảy ra \(\Leftrightarrow y=\frac{59}{14};x=-\frac{171}{28}\)

P/s : Bài này hơi xấu .....

Đặt \(A=x^2+15y^2+xy+8x+y+2020\)

Ta có: \(A=x^2+x\left(y+8\right)+15y^2+y+2020=\left(x^2+x\left(y+8\right)+\frac{\left(y+8\right)^2}{4}\right)\)\(+\left(15y^2+y-\frac{\left(y+8\right)^2}{4}\right)+2020=\left(x+\frac{y+8}{2}\right)^2+\frac{59y^2-12y-64}{4}+2020\)\(=\left(x+\frac{y+8}{2}\right)^2+\frac{59\left(y-\frac{6}{59}\right)^2-\frac{3812}{59}}{4}+2020\ge\frac{\frac{-3812}{59}}{4}+2020=\frac{118227}{59}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}y-\frac{6}{59}=0\\x=-\frac{y+8}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-239}{59}\\y=\frac{6}{59}\end{cases}}\)

a. Vì \(\left|x-1\right|\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^2\ge0\forall x;y\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^2+2020\ge2020\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=0\\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\y=-2\end{cases}}}\)

Vậy Bmin = 2020 <=> x = 1 và y = - 2

b. Vì \(x^2\ge0\forall x\Rightarrow-x^2\le0\)

\(\Rightarrow-x^2+2019\le2019\)

Dấu "=" xảy ra \(\Leftrightarrow-x^2=0\Leftrightarrow x=0\)

Vậy Pmax = 2019 <=> x = 0

Vì \(\left|y-1\right|\ge0\forall y;\left(t+2\right)^4\ge0\forall t\)

\(\Rightarrow-\left|y-1\right|-\left|t+2\right|^4\le0\forall y;t\)

\(\Rightarrow-\left|y-1\right|-\left|t-2\right|^4+21\le21\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|y-1\right|=0\\\left|t+2\right|^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y-1=0\\t+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}y=1\\t=-2\end{cases}}\)

Vậy Qmax <=> y = 1 và t = 2

Cảm ơn bạn Death Note nha

Áp dụng BĐT Bunyakovsky ta được:

\(\left(x+y\right)\left(\frac{2020}{x}+\frac{1}{2020y}\right)\ge\left(\sqrt{x}\cdot\sqrt{\frac{2020}{x}}+\sqrt{y}\cdot\sqrt{\frac{1}{2020y}}\right)\)

\(=\left(\sqrt{2020}+\sqrt{\frac{1}{2020}}\right)^2=2020+\frac{1}{2020}+2=2022\frac{1}{2020}\)

\(\Leftrightarrow\frac{2021}{2020}\cdot S\ge2022\frac{1}{2020}\)

\(\Rightarrow S\ge2022\frac{1}{2020}\div\frac{2021}{2020}=2021\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{\sqrt{x}}{\sqrt{\frac{2020}{x}}}=\frac{\sqrt{y}}{\sqrt{\frac{1}{2020y}}}\\x+y=\frac{2021}{2020}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020y\\x+y=\frac{2021}{2020}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Vậy Min(S) = 2021 khi \(\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

`|x-1|+2020|x-2|+|x-3|`

`=|x-1|+|3-x|+2020|x-2|`

Áp dụng BĐT `|A|+|B|>=|A+B|`

`=>|x-1|+|3-x|>=|x-1+3-x|=2`

Mà `|x-2|>=0=>2020|x-2|>=0`

`=>|x-1|+2020|x-2|+|x-3|>=2`

Dấu "=" xảy ra khi $\begin{cases}(x-1)(3-x) \ge 0\\x-2=0\\\end{cases}$

`<=>` $\begin{cases}(x-1)(x-3) \le 0\\x=2\\\end{cases}$

`<=>` $\begin{cases}1 \le x \le 3\\x=2\\\end{cases}$

`<=>x=2`