a) \(A=x^2+2x+3=x^2+2x+1+2\)
\(=\left(x+1\right)^2+2\ge2\)
Vậy A luôn dương với mọi x
b) \(B=-x^2+4x-5=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+2^2\right)-1\)
\(=-\left(x-2\right)^2-1\le-1\)
Vậy B luôn âm với mọi x
a)\(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)
Vậy x2 +2x+3 luôn dương.
b)\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\le-1\)
Vậy -x2 +4x-5 luôn luôn âm.
a.x2+ 2x+ 3
=x2+ 2.x.1+ 12- 12+ 3
= (x+1)2 -1+3
= (x+1)2+ 2
Ta có: (x+1)2 ≥0
(x+1)2+ 3≥ 3>0
⇒x2+ 2x+ 3>0 mọi x
Vậy x2+ 2x+3>0 mọi x
b. -x2+ 4x- 5
= - (x2- 4x +5)
= - (x2- 2.x.2+ 22- 22+ 5)
= - ((x- 2)2- 4+ 5)
= - ((x- 2)2+1)
= -(x- 2)2 -1
Ta có: (x-2)2 ≥0
- (x-2)2 ≤0
- (x-2)2 +1≤ 1
⇒ -x2+ 4x- 5 <0 mọi x
Vậy -x2+ 4x- 5 <0 mọi x
