Ta có : x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1 \(\ge1\forall x\)

Vậy  x² + 2x + 2 \(>0\forall x\)

Ta có : x² + 2x + 2

=> x² + 2x + 1 + 1

=> ( x + 1)² + 1  >  1\(\forall x\)

Vậy x² + 2x + 2   > \(0\forall x\)

a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x^2-x+1>0\forall x\)

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ

1) \(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

2) \(B=x^2+6x+11=\left(x+3\right)^2+2\ge2>0\left(\forall x\right)\)

3) \(C=4x^2+4x-2=\left(2x+1\right)^2-2\ge-2\) chưa chắc nhỏ hơn 0

4) \(D=-x^2-6x-11=-\left(x+3\right)^2-2\le-2< 0\left(\forall x\right)\)

5) \(E=-4x^2+4x-2=-\left(2x-1\right)^2-1\le-1< 0\left(\forall x\right)\)

1. \(A=x^2+2x+2=\left(x+1\right)^2+1\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(x+1\right)^2+1\ge1\)

=> Đpcm

2. \(B=x^2+6x+11=\left(x+3\right)^2+2\)

Vì \(\left(x+3\right)^2\ge0\forall x\)\(\Rightarrow\left(x+3\right)^2+2\ge2\)

=> Đpcm

3. \(C=4x^2+4x-2=-\left(4x^2-4x+2\right)\)

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow4\left(x-\frac{1}{2}\right)^2+1\ge1\)

\(\Rightarrow-\left(4\left(x-\frac{1}{2}\right)^2+1\right)\le1\)

=> Đpcm

4,5 làm tương tự

\(1.A=x^2+2x+2=\left(x+1\right)^2+1\)

Vì\(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\forall x\)

hay\(\left(x+1\right)^2+1>0\forall x\)

\(2.B=x^2+2x.3+9+2=\left(x+3\right)^2+2\)

CM tương tự A

\(3.C=4x^2+4x-2=\left(2x+1\right)^2-2\)

Vì\(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2-2\ge-2\forall x\)(có thể >0)

4,5 Cm tương tự

a) Xét \(x^2-4x+4=\left(x-2\right)^2\ge0\)

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

= \(\left(x^2-4x+4\right)+\left(x^2-4xy+4y^2\right)+1\)

= \(\left(x-2\right)^2+\left(x-2y\right)^2+1\ge1>0\)

5x^2+2y^2+4xy-4x-y+5=(4x^2+y^2+4xy)+(x^2-4x+4)+(y^2-y+1/4)+3/4 =(2x+y)^2+(x-2)^2+(y-1/2)^2+3/4  (1)

 vi (2x+y)^2>=0 , (x-2)^2>=0  ,(y-1/2)^2>=0 (2)

tu 1 va 2 suy ra dieu phai chung minh

(x+y)^2+(x+2)^2-(-x-y)^2+x^2+y^2+1>=0

a ) \(x^2-4x+5\)

\(=\left(x^2-2.2x+4\right)+1\)

\(=\left(x-2\right)^2+1\\\)

Ta có : \(\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\) ( ĐPCM )

b ) \(x^2-4xy+5y^2\)

\(=\left(x^2-4xy+4y^2\right)+y^2\)

Ta có : \(\left(x-2y\right)^2\ge0\)

\(y^2\ge0\)

\(\Rightarrow\left(x-2y\right)^2+y^2\ge0\) ( ĐPCM )