a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

x²-2x+2=(x²-2x+1)+1=( x-1)²+1

Mà (x-1)²≥0 với mọi x

=> (x-1)²+1>0 với mọi x

=> x²-2x+2>0 với mọi x

x^2 + 2x + 2 = x^2 + 2.x.1 + 1^2 +1 = (x + 1)^2 + 1 > 0

-x^2 + 4x - 4 = -(x^2 - 2.x.2 + 2^2) = -(x - 2)^2 <= 0

a) ta co ; x^2+ 2x+ 2= (x²+2x+1)+1=(x+1)²+1>0

vi (x+1)²>hoặc=0;1>0suy ra x^2+ 2x+ 2>0

b)ta co  -x²+4x-4=-(x²-4x+4)=-(x-2)²<0

a) x^2 + 2x + 2 = ( x^2 + 2x +1 ) + 1 =( x + 1)^2 +1 >0 với mọi x 

b) -x^2 + 4x - 4 = -( x^2 -4x + 4 ) = - ( x - 2)^2  ≤ 0 với mọi x

\(x^2-2\sqrt{2}x+2=\left(x-\sqrt{2}\right)^2\ge0\)

\(x^2-2\sqrt{2}x+2=x^2-2\sqrt{2}x+\left(\sqrt{2}\right)^2=\left(x-\sqrt{2}\right)^2\)

vì \(\left(x-\sqrt{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(x^2-2\sqrt{2}x+2\ge0\forall x\)

Lời giải:

Do $x\geq 2$ nên:

$x-2\geq 0$

$2x-1\geq 2.2-1>0$

Do đó: $(x-2)(2x-1)\geq 0$ (đpcm)

 = (x2-x+1)(x2+3x+10)+10 = P

x²-x+1=(x-\(\frac{1}{2}\))²+\(\frac{3}{4}\)>0

x²+3x+10=(x+\(\frac{3}{2}\))²+\(\frac{31}{4}\)>0

vây P>0

2x^2+4x+3=2(x^2+2x+1)+1=2(x+1)^2+1>0 với mọi x

2x²+4x+3=2(x²+2x+3/2)=2(x²+2x+1+1/2)=2(x+1)²+1>0 với mọi x