Đáp án:
$min_A=0 \Leftrightarrow x=-\dfrac{\sqrt3}{3}\\ max_B= 1+\dfrac{\sqrt{2}}{8} \Leftrightarrow x=\dfrac{1}{2\sqrt{2}}\\ max_C= \dfrac{29}{8}\Leftrightarrow x=\dfrac{\sqrt{5}}{4}\\ min_D= \dfrac{8-9\sqrt{2}}{8} \Leftrightarrow x=\dfrac{3}{2\sqrt{2}}$
Giải thích các bước giải:
$A=2\sqrt{3}x+3x^2+1\\ =3x^2+2\sqrt{3}x+1\\ =(\sqrt{3}x)^2+2\sqrt{3}x+1\\ =(\sqrt{3}x+1)^2 \ge 0 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{3}x+1=0 \Leftrightarrow x=-\dfrac{\sqrt3}{3}$
$B=x-\sqrt{2}x^2+1\\ =-\sqrt{2}x^2+x+1\\ =-(\sqrt[4]{2}x)^2+2.\sqrt[4]{2}x.\dfrac{1}{2\sqrt[4]{2}}-\dfrac{\sqrt{2}}{8}+1+\dfrac{\sqrt{2}}{8}\\ =-\left(\sqrt[4]{2}x-\dfrac{1}{2\sqrt[4]{2}}\right)^2+1+\dfrac{\sqrt{2}}{8} \le 1+\dfrac{\sqrt{2}}{8}$
Dấu "=" xảy ra $\Leftrightarrow \sqrt[4]{2}x-\dfrac{1}{2\sqrt[4]{2}}=0 \Leftrightarrow x=\dfrac{1}{2\sqrt{2}}$
$C=\sqrt{5}x-2x^2+3\\ =-2x^2+\sqrt{5}x+3\\ =-(\sqrt{2}x)^2+2.\sqrt{2}x.\dfrac{\sqrt{5}}{2\sqrt{2}}-\dfrac{5}{8}+\dfrac{29}{8}\\ =-\left(\sqrt{2}x-\dfrac{\sqrt{5}}{2\sqrt{2}}\right)^2+\dfrac{29}{8} \le \dfrac{29}{8} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{2}x-\dfrac{\sqrt{5}}{2\sqrt{2}}\Leftrightarrow x=\dfrac{\sqrt{5}}{4}$
$D=1-3x+\sqrt{2}x^2\\ =\sqrt{2}x^2-3x+1\\ =(\sqrt[4]{2}x)^2-2.\sqrt[4]{2}x.\dfrac{3}{2\sqrt[4]{2}}+\dfrac{9\sqrt{2}}{8}+1-\dfrac{9\sqrt{2}}{8}\\ =-\left(\sqrt[4]{2}x-\dfrac{3}{2\sqrt[4]{2}}\right)^2+\dfrac{8-9\sqrt{2}}{8} \le \dfrac{8-9\sqrt{2}}{8}$
Dấu "=" xảy ra $\Leftrightarrow \sqrt[4]{2}x-\dfrac{3}{2\sqrt[4]{2}}=0 \Leftrightarrow x=\dfrac{3}{2\sqrt{2}}$
