Đáp án:

$12)\\ a)min_M=\dfrac{31}{4} \Leftrightarrow x=\dfrac{3}{2}\\ b)min_N=-120 \left\{\begin{array}{l} x=-4\\ y=2\end{array} \right.\\ 13)\\ a)max_A=12 \Leftrightarrow a=\dfrac{3}{2}\\ b)max_B=1 \Leftrightarrow m=2$

Giải thích các bước giải:

$12)\\ a)M=x^2-3x+10\\ =x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4} \ge \dfrac{31}{4} \ \forall x$

Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{3}{2}=0 \Leftrightarrow x=\dfrac{3}{2}$ $b)N=2x^2+5y^2+4xy+8x-4y-100\\ =2(x^2+2xy+y^2)+3y^2+8x-4y-100\\ =2(x+y)^2+8x+8y+8 +3y^2-12y+12-120\\ =2\left((x+y)^2+2(x+y).2+4\right)+3(y^2-4y+4)-120\\ =2(x+y+2)^2+3(y-2)^2-120 \ge -120 \ \forall \ x,y$

Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} x+y+2=0\\ y-2=0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x=-4\\ y=2\end{array} \right.$

$13)\\ a)A=12a-4a^2+3\\ =-4a^2+2.2a.3-9+12\\ =-(2a-3)^2+12 \le 12 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow 2a-3=0 \Leftrightarrow a=\dfrac{3}{2}$

$b)B=m-\dfrac{m^2}{4}\\ =-\dfrac{m^2}{4}+2.\dfrac{m}{2}.1-1+1\\ =-\left(\dfrac{m}{2}-1\right)^2+1  \le 1 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow \dfrac{m}{2}-1=0 \Leftrightarrow m=2$