`***` Lời giải chi tiết `***`

`1)`

`x^{2}-6x+11`

`=(x^{2}-6x+9)+2`

`=(x-3)^{2}+2>=2` với mọi `x`

Dấu `=` xảy ra `<=>x-3=0<=>x=3`

Vậy `min=2<=>x=3`

`2)`

`5x-x^{2}`

`=-(x^{2}-5x)`

`=-(x^{2}-5x+(25)/(4)-(25)/(4))`

`=-(x-(5)/(2))^{2}+(25)/(4)≤(25)/(4)` với mọi `x`

Dấu `=` xảy ra `<=>x-(5)/(2)=0<=>x=(5)/(2)`

Vậy `max=(25)/(4)<=>x=(5)/(2)`

`3)`

`5-8x-x^{2}`

`=-(x^{2}+8x-5)`

`=-(x^{2}+8x+16-21)`

`=-(x+4)^{2}+21≤21` với mọi `x`

Dấu `=` xảy ra `<=>x+4=0<=>x=-4`

Vậy `max=21<=>x=-4`

`4)`

`4x-x^{2}+3`

`=-(x^{2}-4x-3)`

`=-(x^{2}-4x+4-7)`

`=-(x-2)^{2}+7≤7` với mọi `x`

Dấu `=` xảy ra `<=>x-2=0<=>x=2`

Vậy `max=7<=>x=2`

`5)`

`x^{2}-2x+y^{2}+4y+8`

`=(x^{2}-2x+1)+(y^{2}+4y+4)+3`

`=(x-1)^{2}+(y+2)^{2}+3≥3` với mọi `x;y`

Dấu `=` xảy ra `<=>x=1;y=-2`

Vậy `min=3<=>x=1;y=-2`

`6)`

`x^{2}-4x+y^{2}-8y+6`

`=(x^{2}-4x+4)+(y^{2}-8y+16)-14`

`=(x-2)^{2}+(y-4)^{2}-14≥ -14` với mọi `x;y`

Dấu `=` xảy ra `<=>x=2;y=4`

Vậy `min=-14<=>x=2;y=4`

`~rai~`

\(1)x^2-6x+11\\\quad\quad=(x^2-6x+9)+2\\\quad\quad=(x-3)^2+2\\\text{Ta có:}(x-3)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow (x-3)^2+2\ge 2\quad\forall x\in\mathbb{R}\\\text{Dấu "=" xảy ra}\Leftrightarrow (x-3)^2=0\Leftrightarrow x=3.\\2)5x-x^2\\=\dfrac{25}{4}-\left(x^2-5x+\dfrac{25}{4}\right)\\=\dfrac{25}{4}-\left(x-\dfrac{5}{2}\right)^2\\\text{Ta có:}\left(x-\dfrac{5}{2}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -\left(x-\dfrac{5}{2}\right)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow \dfrac{25}{4}-\left(x-\dfrac{5}{2}\right)^2\le \dfrac{25}{4}\quad\forall x\in\mathbb{R}\\\text{Dấu "=" xảy ra}\Leftrightarrow \left(x-\dfrac{5}{2}\right)^2=0\\\Leftrightarrow x=\dfrac{5}{2}.\\3)5-8x-x^2\\=21-(x^2+8x+16)\\=21-(x+4)^2\\\text{Ta có:}(x+4)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -(x+4)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 21-(x+4)^2\le 21\quad\forall x\in\mathbb{R}\\\text{Dấu "=" xảy ra}\Leftrightarrow (x+4)^2=0\Leftrightarrow x=-4.\\4)4x-x^2+3\\=7-(x^2-4x+4)\\=7-(x-2)^2\\\text{Ta có:}(x-2)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -(x-2)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 7-(x-2)^2\le 7\quad\forall x\in\mathbb{R}\\\text{Dấu "=" xảy ra}\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2.\\5)x^2-2x+y^2+4y+8\\=(x^2-2x+1)+(y^2+4y+4)+3\\=(x-1)^2+(y+2)^2+3\\\text{Ta có:}(x-1)^2\ge 0;(y+2)^2\ge 0\quad\forall x;y\in\mathbb{R}.\\\Leftrightarrow (x-1)^2+(y+2)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-1)^2+(y+2)^2+3\ge 3\quad\forall x;y\in\mathbb{R}.\\\text{Dấu "=" xảy ra}\Leftrightarrow \begin{cases}(x-1)^2=0\\(y+2)^2=0\end{cases}\Leftrightarrow \begin{cases}x=1\\y=-2.\end{cases}\\6)x^2-4x+y^2-8y+6\\=(x^2-4x+4)+(y^2-8y+16)-14\\=(x-2)^2+(y-4)^2-14\\\text{Ta có:}(x-2)^2\ge 0;(y-4)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-2)^2+(y-4)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-2)^2+(y-4)^2-14\ge -14\quad\forall x;y\in\mathbb{R}\\\text{Dấu "=" xảy ra}\Leftrightarrow \begin{cases}(x-2)^2=0\\(y-4)^2=0\end{cases}\Leftrightarrow \begin{cases}x=2\\y=4.\end{cases}\)