Đáp án:
$10)\\ a)min_M= \dfrac{31}{4} \Leftrightarrow x=\dfrac{3}{2}\\ b)min_N=-120 \Leftrightarrow \left\{\begin{array}{l} x=-4 \\ y=2\end{array} \right.$
Giải thích các bước giải:
$9)\\ a)a^2-2a+2 =a^2-2a+1 +1 =(a-1)^2+1>0 \ \forall \ a\\ b)6b-b^2-10\\ =-b^2+6b-9-1\\ =-(b-3)^2-1<0 \ \forall \ b\\ 10)\\ a)M=x^2-3x+10\\ =x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{31}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4} \ge \dfrac{31}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{3}{2} =0 \Leftrightarrow x=\dfrac{3}{2}$
$b)N=2x^2+5y^2+4xy+8x-4y-100\\ =x^2+4xy+4y^2+x^2+8x+16+y^2-4y+4-120\\ =(x+2y)^2+(x+4)^2+(y-2)^2-120 \ge -120 \ \forall \ x,y$
Dấu "=" xảy ra $\left\{\begin{array}{l} x+2y=0\\ x+4=0\\ y-2=0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x=-4 \\ y=2\end{array} \right.$
