$\text{Đáp án + Giải thích các bước giải:}$

`A=1+6x-x^{2}`

`=>A=-(x^{2}-6x-1)`

`=>A=-(x^{2}-6x+9)+10`

`=>A=-(x-3)^{2}+10`

`\text{Vì}` `(x-3)^{2}≥0`

`=>-(x-3)^{2}≤0`

`=>-(x-3)^{2}+10≤10`

`\text{Dấu = xảy ra khi:}`

`(x-3)^{2}=0`

`=>x=3`

`\text{Vậy GTLN của A là : 10 khi x = 3}`

`---------`

`B=-2x^{2}+6x+8`

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

`=>B=-2(x^{2}-3x+(9)/(4))+(25)/(2)`

`=>B=-2(x-(3)/(2))^{2}+(25)/(2)`

`\text{Vì}` `(x-(3)/(2))^{2}≥0`

`=>-2(x-(3)/(2))^{2}≤0`

`=>-2(x-(3)/(2))^{2}+(25)/(2)≤(25)/(2)`

`\text{Dấu = xảy ra khi :}`

`(x-(3)/(2))^{2}=0`

`=>x=(3)/(2)`

`\text{Vậy GTLN của B là}` `(25)/(2)` `\text{khi x =}` `(3)/(2)`

`---------------`

`C=x^{2}+3y^{2}-2xy-2y`

`=>C=(x^{2}-2xy+y^{2})+(2y^{2}-2y)`

`=>C=(x-y)^{2}+2(y^{2}-y+(1)/(4))-(1)/(2)`

`=>C=(x-y)^{2}+2(y-(1)/(2))^{2}-(1)/(2)`

`\text{Vì}` `(x-y)^{2}≥0;2(y-(1)/(2))^{2}≥0`

`=>(x-y)^{2}+2(y-(1)/(2))^{2}≥0`

`=>(x-y)^{2}+2(y-(1)/(2))^{2}-(1)/(2)≥-(1)/(2)`

`\text{Dấu = xảy ra khi :}`

$\left\{\begin{matrix}(x-y)^{2}=0& \\(y-\dfrac{1}{2})^{2}=0& \end{matrix}\right.$

`=>` $\left\{\begin{matrix}x-y=0& \\y-\dfrac{1}{2}=0& \end{matrix}\right.$

`=>` $\left\{\begin{matrix}x=y& \\y=\dfrac{1}{2}& \end{matrix}\right.$

`=>` $\left\{\begin{matrix}x=\dfrac{1}{2}& \\y=\dfrac{1}{2}& \end{matrix}\right.$

`\text{Vậy GTNN của C là :}` `-(1)/(2)` `\text{khi x = y =}` `(1)/(2)`

Đáp án + Giải thích các bước giải:

$\rm a) \\ A=1+6x-x^2 \\ =-(x^2-6x-1) \\ =-(x^2-6x+9)+10\\ =-(x-3)^2+10 \\ Ta \ có \ : \ (x-3)^2 \geq 0 \to -(x-3)^2 \leq 0 \\ \to -(x-3)^2+10 \geq 10 \\ Dấu \ "=" \ xảy \ ra \ khi \ x-3=0 \\ \to x=3 \\ Vậy \ A_{max}=10 \ khi \ x=3 \\ b) \\ B=-2x^2+6x+8\\=-2(x^2-3x-4)\\=-2(x^2-2 . x . \dfrac{3}{2}+\dfrac{9}{4}) - \dfrac{9}{4}-4) \\ =-2[(x-\dfrac{3}{2})^2-\dfrac{25}{4}]\\=-2(x-\dfrac{3}{2})^2+\dfrac{25}{2} \\ Ta \ có \ : \ (x-\dfrac{3}{2})^2 \geq 0 \\ \to -2(x-\dfrac{3}{2}) \leq 0 \\ \to -2(x-\dfrac{3}{2})^2+\dfrac{25}{2} \leq \dfrac{25}{2} \\ Dấu \ "=" \ xảy \ ra \ khi \ x-\dfrac{3}{2}=0 \\ \to x=\dfrac{3}{2} \\ Vậy \ B_{max}=\dfrac{25}{2} \ khi \ x=\dfrac{3}{2} \\ c) \\ C=x^2+3y^2-2xy-2y \\ =(x^2-2xy+y^2)+(2y^2-2y) \\ =(x-y)^2+2(y^2-y) \\ =(x-y)^2+2[(y^2-2 . y . \dfrac{1}{2} + \dfrac{1}{4})-\dfrac{1}{4}] \\ =(x-y)^2+2[(y-\dfrac{1}{2})^2-\dfrac{1}{4}] \\ =(x-y)^2+2(y-\dfrac{1}{2})^2-\dfrac{1}{2} \\ Ta \ có \ : \ (x-y)^2 \geq 0 ; 2(y-\dfrac{1}{2})^2 \geq 0 \\ \to (x-y)^2+2(y-\dfrac{1}{2})^2 \geq 0 \\ \to (x-y)^2+2(y-\dfrac{1}{2})^2-\dfrac{1}{2} \geq -\dfrac{1}{2} \\ Dấu \ "=" \ xảy \ ra \ khi \ (x-y)^2=0 \ ; \ (y-\dfrac{1}{2})^2=0 \\ \to x=y=\dfrac{1}{2} \\ Vậy \ C_{min}=-\dfrac{1}{2} \ khi \ x=y=\dfrac{1}{2}$