1. $A\,=-2x^2-3x+7\\\quad =-2\left(x^2+\dfrac{3}{2}x-\dfrac{7}{2}\right)\\\quad =-2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{65}{16}\right)\\\quad =-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{65}{8}$
Nhận thấy: $\left(x+\dfrac{3}{4}\right)^2\ge 0$
$↔-2\left(x+\dfrac{3}{4}\right)^2\le 0\\↔-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{65}{8}\le \dfrac{65}{8}\\→\max A=\dfrac{65}{8}$
$→$ Dấu "=" xảy ra khi $x+\dfrac{3}{4}=0$
$↔x=-\dfrac{3}{4}$
Vậy $\max A=\dfrac{65}{8}$ khi $x=-\dfrac{3}{4}$
2. $B\,=x-x^2\\\quad =-x^2+x\\\quad =-x^2+2.x.\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\\\quad =-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{4}\\\quad =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}$
Nhận thấy: $\left(x-\dfrac{1}{2}\right)^2\ge 0$
$↔-\left(x-\dfrac{1}{2}\right)^2\le 0\\↔B\le \dfrac{1}{4}\\→\max B=\dfrac{1}{4}$
$→$ Dấu "=" xảy ra khi $x-\dfrac{1}{2}=0$
$↔x=\dfrac{1}{2}$
Vậy $\max B=\dfrac{1}{4}$ khi $x=\dfrac{1}{2}$
3. $C\,=-4x^2-y^2-4x+8y+10\\\quad =(-4x^2-4x-1)+(-y^2+8y-16)+27\\\quad =-(4x^2+4x+1)-(y^2-8y+16)+27\\\quad =-(2x+1)^2-(y-4)^2+27$
Nhận thấy: $\begin{cases}(2x+1)^2\ge 0\\(y-4)^2\ge 0\end{cases}$
$↔\begin{cases}-(2x+1)^2\le 0\\-(y-4)^2\le 0\end{cases}\\→-(2x+1)^2-(y-4)^2\le 0\\↔C\le 27\\→\max B=27$
$→$ Dấu "=" xảy ra khi $\begin{cases}2x+1=0\\y-4=0\end{cases}$
$↔\begin{cases}2x=-1\\y=4\end{cases}\\↔\begin{cases}x=-\dfrac{1}{2}\\y=4\end{cases}$
Vậy $\max C=27$ khi $x=-\dfrac{1}{2},y=4$
4. $F\,=12x-4y-9x^2-y^2+1\\\quad =(-9x^2+12x-4)+(-y^2-4y-4)+9\\\quad =-(9x^2-12x+4)-(y^2+4y+4)+9\\\quad =-(3x-2)^2-(y+2)^2+9$
Nhận thấy: $\begin{cases}(3x-2)^2\ge 0\\(y+2)^2\ge 0\end{cases}$
$↔\begin{cases}-(3x-2)^2\le 0\\-(y+2)^2\le 0\end{cases}\\→-(3x-2)^2-(y+2)^2\le 0\\↔F\le 9\\→\max F=9$
$→$ Dấu "=" xảy ra khi $\begin{cases}3x-2=0\\y+2=0\end{cases}$
$↔\begin{cases}3x=2\\y=-2\end{cases}\\↔\begin{cases}x=\dfrac{2}{3}\\y=-2\end{cases}$
Vậy $\max F=9$ khi $x=\dfrac{2}{3},y=-2$
