Đáp án:
$a) min_A= -33 \Leftrightarrow x=-6\\ b)min_B= \dfrac{1}{4} \Leftrightarrow x=-\dfrac{1}{2}\\ c)min_C=\dfrac{1}{3} \Leftrightarrow x=y=\dfrac{1}{3}\\ d)min_D= 1984 \Leftrightarrow x=0;x=-5$
Giải thích các bước giải:
$a) A= x^2 + 12x + 3\\ = x^2 + 2.x.6 + 36-33\\ =(x+6)^2-33 \ge -33 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x+6=0 \Leftrightarrow x=-6$
$b)B= 4x^4 + 2x + 1\\ = 4x^4+4x^3+x^2-4x^3-4x^2-x+3x^2+3x+\dfrac{3}{4}+\dfrac{1}{4}\\ =x^2(4x^2+4x+1)-x(4x^2+4x+1)+\dfrac{3}{4}\left(4x^2+4x+1\right)+\dfrac{1}{4}\\ =x^2(2x+1)^2-x(2x+1)^2+\dfrac{3}{4}(2x+1)^2+\dfrac{1}{4}\\ =(2x+1)^2\left(x^2-x+\dfrac{3}{4}\right)+\dfrac{1}{4}\\ =(2x+1)^2\left(x^2-x+\dfrac{1}{4}+\dfrac{1}{2}\right)+\dfrac{1}{4}\\ =(2x+1)^2\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\right)+\dfrac{1}{4}\ge \dfrac{1}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 2x+1=0 \Leftrightarrow x=-\dfrac{1}{2}$
$c)x + 2y = 1 \Rightarrow x=1-2y\\ C=(1-2y)^2+2y^2\\ =6y^2-4y+1\\ =(\sqrt{6}y)^2-2.\sqrt{6}y.\dfrac{\sqrt{6}}{3}+\dfrac{2}{3}+\dfrac{1}{3}\\ =\left(\sqrt{6}y-\dfrac{\sqrt{6}}{3}\right)^2+\dfrac{1}{3} \ge \dfrac{1}{3} \ \forall \ y$
Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} x=1-2y\\ \sqrt{6}y-\dfrac{\sqrt{6}}{3} \end{array} \right. \Leftrightarrow x=y=\dfrac{1}{3}$
$d)D= (x-1)(x+2)(x+3)(x+6) + 2020\\ = (x-1)(x+6)(x+2)(x+3) + 2020\\ = (x^2+5x-6)(x^2+5x+6) + 2020(1)\\ t=x^2+5x\\ (1) \Leftrightarrow (t-6)(t+6)+ 2020\\ =t^2-36+ 2020\\ =t^2+1984 \ge 1984 \ \forall \ t$
Dấu "=" xảy ra $\Leftrightarrow t=0 \Leftrightarrow x^2+5x=0\Leftrightarrow x(x+5)=0\Leftrightarrow x=0;x=-5$
