Đáp án:
$x^2+3x+7>0$
$=x^2+2.x.\dfrac{3}{2} +\dfrac{9}{4} +\dfrac{19}{4} >0$
$=(x+\dfrac{3}{2})^2 +\dfrac{19}{4} >0$
Vì $(x+\dfrac{3}{2})^2 ≥ 0$
Nên $(x+\dfrac{3}{2})^2 +\dfrac{19}{4} > 0$
$5x-x^2-8 <0$
$=-(x^2-5x+8) < 0$
$=-(x^2 -2 .x.\dfrac{5}{2} +\dfrac{25}{4} +\dfrac{7}{4} )< 0$
$=-(x+\dfrac{5}{2})^2 -\dfrac{7}{4} < 0$
Vì $-(x+\dfrac{5}{2})^2 ≤ 0$
Nên $-(x+\dfrac{5}{2})^2 -\dfrac{7}{4} < 0$
