a) $x^2+x+1=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}$
$=(x+\dfrac{1}{2})^2+\dfrac{3}{4}$
Ta thấy: $(x+\dfrac{1}{2})^2≥0$
$→$ Dấu "=" xảy ra khi $x+\dfrac{1}{2}=0$
$→x=-\dfrac{1}{2}$
$→A_{min}=\dfrac{3}{4}$ khi $x=-\dfrac{1}{2}$
b) $-x^2+x+1=-(x^2-x-1)$
$=-(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4})$
$=-(x-\dfrac{1}{2})^2+\dfrac{5}{4}$
Ta thấy: $-(x-\dfrac{1}{2})^2≤0$
$→-(x-\dfrac{1}{2})^2+\dfrac{5}{4}≤\dfrac{5}{4}$
$→$ Dấu "=" xảy ra khi $x-\dfrac{1}{2}=0$
$→x=\dfrac{1}{2}$
$→B_{max}=\dfrac{5}{4}$ khi $x=\dfrac{1}{2}$
c) $-4x^2-6x=-(4x^2-6x)$
$=-(4x^2+6x+\dfrac{9}{4}-\dfrac{9}{4})$
$=-(4x^2+6x+\dfrac{9}{4})+\dfrac{9}{4}$
$=-(2x+\dfrac{3}{2})^2+\dfrac{9}{4}$
Ta thấy: $-(2x+\dfrac{3}{2})^2≤0$
$→-(2x+\dfrac{3}{2})+\dfrac{9}{4}≤9$
$→$ Dấu "=" xảy ra khi $2x+\dfrac{3}{2}=0$
$→x=-\dfrac{3}{4}$
$→C_{max}=\dfrac{9}{4}$ khi $x=-\dfrac{3}{4}$
