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) x^2 + x +1 = x^2 + 1/2x+1/2x + 1/4 + 3/4= x(x+1/2)+1/2(x+1/2) + 3/4

=( x+1/2)^2 + 3/4

Do (x+1/2)^2 lớn hơn hoặc  = 0 vs mọi x => (x+1/2)^2 + 3/4 >0 =>  x^2 + x +1 > 0 với mọi x

Câu a :

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

Vậy biểu thức trên luôn lớn hơn 0 với mọi x

Làm Full cho you nhé,bạn kia sai r:

\(linh_1=x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(đpcm\right)\)

\(linh_2=-4x^2-4x-2=-1\left(4x^2+4x+2\right)=-1\left(4x^2+4x+1+1\right)=-1\left(4x^2+4x+1\right)-1=-1\left(2x+1\right)^2-1< 0\left(đpcm\right)\)

Bài làm:

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

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

=> đpcm

b) \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

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

    Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)

              \(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)

              \(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )

b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

        \(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

        \(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)

    Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )

a) Ta có : -4x² - 4x - 2 = -(4x² + 4x + 1) - 1 = -(2x + 1)² - 1 < 0 (đpcm)

b) x² + 4y² + z² - 2x - 6z + 8y + 15

= (x² - 2x + 1) + (z² - 6z + 9) + (4y² + 8y + 4) + 1

= (x - 1)² + (z - 3)² + 4(y + 1)² + 1 > 0 (đpcm)

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

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

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

Vì \(\left(2x+1\right)^2\ge0\Rightarrow-\left(2x+1\right)^2\le0\Rightarrow-\left(2x+1\right)^2-1\le-1< 0\)

Vậy ta có đpcm.

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

Ta có \(\left(2x+1^2\right)\ge0\)\(=>-[\left(2x+1\right)^2+1]\le-1< 0\)

Vậy ... 

 Học tốt nha !

2x^2+4x+2+1>0

2(x+1)^2+1>0 (đúng) 

suy ra đpcm 

\(2x^2+4x+3\)

\(=2\left(x^2+2x+\frac{3}{2}\right)\)

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

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

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