Đáp án:
$3) max_A=-\dfrac{2}{3} x=\dfrac{1}{3}\\ 4)\max_A= 1 \Leftrightarrow x=-4\\ 5) max_A= -6 \ \Leftrightarrow x=3\\ 7)max_A= -\dfrac{3}{4} \Leftrightarrow x=-\dfrac{1}{2}\\ 8) max_A= -\dfrac{7}{2} \Leftrightarrow x=\dfrac{3}{2}\\ 9)max_A= \dfrac{25}{4} \Leftrightarrow x=\dfrac{5}{2}\\ 10) max_A= \dfrac{1}{16} \Leftrightarrow x=-\dfrac{1}{8}$
Giải thích các bước giải:
$3)\\ A=-3x^2+2x-1\\ =-(3x^2-2x+1)\\ =-\left((\sqrt{3}x)^2-2.\sqrt{3}x.\dfrac{1}{\sqrt{3}}+\left(\dfrac{1}{\sqrt{3}}\right)^2+\dfrac{2}{3}\right)\\ =-\left(\sqrt{3}x-\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{2}{3} \le -\dfrac{2}{3} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{3}x-\dfrac{1}{\sqrt{3}}=0 \Leftrightarrow x=\dfrac{1}{3}$
$4)\\ A=-\dfrac{1}{4}x^2-2x-3\\ =-\left(\dfrac{1}{4}x^2+2x+3\right)\\ =-\left(\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.2+2^2-1\right)\\ =-\left(\dfrac{1}{2}x+2\right)^2+1 \le 1 \ \forall \ x$
Dấu "=" xảy ra $ \Leftrightarrow \dfrac{1}{2}x+2=0 \Leftrightarrow x=-4$
$5)\\ A=-4x^2+12x-7\\ =-(4x^2-12x+7)\\ =-\left((2x)^2-2.2x.3+3^2-2\right)\\ =-(2x-3)^2+2 \le 2 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 2x-3=0 \Leftrightarrow x=\dfrac{3}{2}$
$6)\\ A=-x^2+6x-15\\ =-(x^2-6x+15)\\ =-(x^2-6x+9+6)\\ =-(x^2-6x+9)-6\\ =-(x-3)^2-6 \le -6 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-3=0 \Leftrightarrow x=3$
$7)\\ A=-x^2-x-1\\ =-(x^2+x+1)\\ =-\left(x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{3}{4}\\ =-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4} \le -\dfrac{3}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x+\dfrac{1}{2}=0 \Leftrightarrow x=-\dfrac{1}{2}$
$8)\\ A=-2x^2+6x-8\\ =-2(x^2-3x+4)\\ =-2\left(x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\right)\\ =-2\left(x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right)-\dfrac{7}{2}\\ =-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{2} \le -\dfrac{7}{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{3}{2}=0 \Leftrightarrow x=\dfrac{3}{2}$
$9)\\ A=-x^2+5x\\ =-(x^2-5x)\\ =-\left(x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\right)\\ =-\left(x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right)+\dfrac{25}{4}\\ =-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4} \le \dfrac{25}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{5}{2}=0 \Leftrightarrow x=\dfrac{5}{2}$
$10)\\ A=-4x^2-x\\ =-(4x^2+x)\\ =-\left((2x)^2+2.2x.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^2\right)\\ =-\left((2x)^2+2.2x.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right)+\dfrac{1}{16}\\ =-\left(2x+\dfrac{1}{4}\right)^2+\dfrac{1}{16} \le \dfrac{1}{16} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 2x+\dfrac{1}{4}=0 \Leftrightarrow x=-\dfrac{1}{8}$
