Đáp án:
$a)min_A= 2 \Leftrightarrow x=1\\ b)min_B= -\dfrac{49}{12} \Leftrightarrow x=-\dfrac{5}{6}\\ c)min_C= 1 x=y=2\\ d)min_D= -2 \Leftrightarrow \left\{\begin{array}{l} x=1\\ y=-1 \end{array} \right.$
Giải thích các bước giải:
$a)A=x^2-2x+3\\ =x^2-2x+1+2\\ =(x-1)^2+2 \ge 2 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-1=0 \Leftrightarrow x=1$
$b)B=3x^2+5x-2\\ =(\sqrt{3}x)^2+2.\sqrt{3}x.\dfrac{5\sqrt{3}}{6}+\dfrac{25}{12}-\dfrac{49}{12}\\ =\left(\sqrt{3}x+\dfrac{5\sqrt{3}}{6}\right)^2-\dfrac{49}{12} \ge -\dfrac{49}{12} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{3}x+\dfrac{5\sqrt{3}}{6}=0 \Leftrightarrow x=-\dfrac{5}{6}$
$c)x^2+2y^2-2xy-4y+5\\ =x^2-2xy+y^2+y^2-4y+4+1\\ =(x-y)^2+(y-2)^2+1 \ge 1 \ \forall \ x,y$
Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} x-y=0\\ y-2=0\end{array} \right. \Leftrightarrow x=y=2$
$d)D=5x^2+8xy+5y^2-2x+2y\\ =4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1-2\\ =(2x+2y)^2+(x-1)^2+(y+1)^2-2 \ge -2 \ \forall \ x,y$
Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} 2x+2y=0\\ x-1=0 \\ y-1 =0 \end{array} \right. \left\{\begin{array}{l} x=1\\ y=-1 \end{array} \right.$
