ta có x^2-x+2

  =x^2-2x.1/2+(1/2)^2-(1/2)^2+2

  =(x-1/2)^2+7/4

ta có (x-1/2)^2 lớn hơn hoặc bằng 0 với mọi x  (1)

          7/4 lớn hơn 0  (2)

từ (1),(2) suy ra (x-1/2)^2+7/4 lớn hơn 0

vậy x^2-x+2 lớn hơn 0 với mọi x

a) Ta có: \(x^2-20x+101=x^2-2.x.10+10^2+1=\left(x-10\right)^2+1\)

Vì \(\left(x-10\right)^2\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow\left(x-10\right)^2+1>1>0\)

Vậy x²-20x+101 >0 với mọi x

b) \(4a^2+4a+2=\left(2a\right)^2+2.2a.1+1+1=\left(2a+1\right)^2+1\)

Vì \(\left(2a+1\right)^2\ge0\left(\forall a\in Z\right)\)

\(\Rightarrow\left(2a+1\right)^2+1>1>0\)

Vậy 4a²+4a+2 > 0 với mọi a

c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+20\right)^2\) \(\ge0\left(\forall x\right)\)

Giúp mình với !!

\(x^2+8x+17=(x^2+2.4x+16)+1=(x+4)^2+1\geq1>0\)

\(\Rightarrow x^2+8x+17 > 0 \) với mọi x

\(\Rightarrow đpcm\)

\(x^2+8x+17=x^2+8x+16+1=\left(x+4\right)^2+1\ge1>0\forall x\)

Hay: \(x^2+8x+17>0\forall x\)

=.= hok tốt!!

\(\left(x^2+2\cdot x\cdot4+4^2\right)-4^2+17\)

\(\rightarrow\left(x+4\right)^2+1\)

MÀ \(\left(x+4\right)^2\ge0\)

\(\Rightarrow\left(x+4\right)^2+1\ge1\) HAY \(x^2+8x+17\ge1\forall x\)

Hok tốt

\(x^2-8x+20=\left(x^2-8x+16\right)+4=\left(x-4\right)^2+4\ge4>0\forall x\)

Bài 1: 

\(2x^2+8x+30\)

\(=2\left(x^2+4x+15\right)\)

\(=2\left(x^2+4x+4+11\right)\)

\(=2\left(x+2\right)^2+22>0\forall x\)

Bài 2: 

\(-x^2-2x-12\)

\(=-\left(x^2+2x+12\right)\)

\(=-\left(x^2+2x+1+11\right)\)

\(=-\left(x+1\right)^2-11< 0\forall x\)

a) Ta có \(2x^2-8x+13=2x^2-8x+8+5\)

\(=2\left(x^2-4x+4\right)+5\)

\(=2\left(x-2\right)^2+5\ge5\forall x\)

Giả sử trước khi làm nhé 

\(a)\)\(2x^2-8x+13>0\)

\(\Leftrightarrow\)\(4x^2-16x+26>0\)

\(\Leftrightarrow\)\(\left(4x^2-16+16\right)+10>0\)

\(\Leftrightarrow\)\(\left(2x-4\right)^2+10\ge10>0\) ( luôn đúng ) 

Vậy ... 

\(b)\)\(-2+2x-x^2< 0\)

\(\Leftrightarrow\)\(x^2-2x+2>0\)

\(\Leftrightarrow\)\(\left(x^2-2x+1\right)+1>0\)

\(\Leftrightarrow\)\(\left(x-1\right)^2+1\ge1>0\) ( luôn đúng ) 

Vậy ... 

Chúc bạn học tốt ~ 

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

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

Do \(-\left(x-1\right)^2\le0\)

\(\Rightarrow-\left(x-1\right)^2-1\le0-1< 0\left(dpcm\right)\)

ta co 0^1=0^2=...=0^n=0

1^1=1^2=...=1^n=1

Ta có : \(0^1=0^3=\cdot\cdot\cdot=0^n=0\left(n\ge2\right)\)

\(1^1=1^2=\cdot\cdot\cdot=1^n=1\left(n\ge2\right)\)

Vậy bài toán đã được chứng minh

ta có \(A=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ge0\)