`C=|x+2|+|x-4|+2020`
`=|x+2|+|4-x|+2020`
Áp dụng BĐT `|A|+|B|>=|A+B|`
`=>|x+2|+|4-x|>=|x+2+4-x|=6`
`=>C>=2020+6=2026`
Dấu "=" xảy ra khi `(x+2)(4-x)>=0<=>(x+2)(x-4)<=0<=>-2<=x<=4`

Áp dụng BĐT 


Dấu "=" xảy ra khi 

Ta có:

\(A=\left|x-4\right|+\left|x-2020\right|=\left|x-4\right|+\left|2020-x\right|\ge x-4+2020-x=2016\)

Dấu "=" xảy ra <=> x - 4 \(\ge0\)

                          và 2020 - x \(\ge0\)

<=> \(x\ge4\) và \(x\le2020\)

\(\Leftrightarrow4\le x\le2020\)

Vậy A đạt GTNN là 2016 \(\Leftrightarrow4\le x\le2020\)

7< y : 4 < 9

Lời giải:

 Ta thấy: $(x-9)^{2020}=[(x-9)^{1010}]^2\geq 0$ với mọi $x$

$(y-3)^{30}=[(y-3)^{15}]^2\geq 0$ với mọi $y$

$\Rightarrow C\geq 0+4.0-25=-25$
Vậy GTNN của $C$ là $-25$. Giá trị này đạt tại $x-9=y-3=0$

$\Rightarrow x=9; y=3$

\(x^4-2x^2+1+x^2+2x+1+2018=\left(x^2-1\right)^2+\left(x+1\right)^2+2018\ge2018\)

Dấu "=" xayr ra <=> \(\hept{\begin{cases}x^2-1=0\\x+1=0\end{cases}\Leftrightarrow x=-1}\)

Kết luận :...

 Ta có : \(\left|x-2019\right|\ge x-2019\). Dấu "=" khi \(x-2019\ge0\)

             \(\left|x-2020\right|=\)\(\left|2020-x\right|\ge2020-x\).Dấu "=" khi \(2020-x\ge0\)

=> \(\left|x-2019\right|+\left|2020-x\right|\)\(\ge x-2019+2020-x\)

=> \(\left|x-2019\right|+\left|x-2020\right|+2\)\(\ge3\)

hay \(A\ge3\)

\(MinA=3\Leftrightarrow\)\(\hept{\begin{cases}x-2019\ge0\\2020-x\ge0\end{cases}}\)\(\Leftrightarrow2019\le x\le2020\)

Ta có : \(\left(x-1\right)^2\ge0\)

            \(\left|y-5\right|\ge0\)

            \(\sqrt{z-4}\ge0\)

Để có được \(Min_A\Leftrightarrow\hept{\begin{cases}x-1=0\\y-5=0\\z-4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=5\\z=4\end{cases}}\)

\(\Leftrightarrow A=1^2+0+0+0+2020=2021\)

Vậy \(Min_A=2021\Leftrightarrow\left(x;y;z\right)=\left(1;5;4\right)\)

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

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

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

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

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

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

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