$a) x^2-x+1>0$
->$ = [x^2-x+(1/2)^2]+3/4$
$=(x-1/2)^2+3/4$
Do $( x-1/2)^2>=0$->$(x-1/2)^2+3/4>=3/4$
Vậy $x^2-x+1>0$ với mọi$x$
$b) x-x^2-2<0$
->=$ -(x^2-x+2)$
=$ -(x^2-x+1/4)-7/4$
=$-(x-1/2)^2-7/4$
Do $-(x-1/2)^2<=0$
-> $-(x-1/2)^2-7/4<=-7/4$
Vậy $ x-x^2-2<0$ với mọi $x$
$#Hoctot$
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