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. Huhu T^T mong sẽ có ai đó giúp mình "((

hơi ngán dạng này :((((

a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

b,

\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)

c,

\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,

\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))

Ta có: \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\)

Thấy \(x^8\ge0;x^5< x^8\Rightarrow x^8-x^5\ge0\)

\(\Rightarrow x^8-x^5+x^2-x+1>0\forall x\in R.\)(đpcm) 

sai rồi bạn ơi

x = 0 

co dung KO thi nho ket ban voi minh nha ^_^

a)Ta có: x²+x+1

=x²+2.x.¹/₂+¹/₄+³/₄

=(x+¹/₂)²+³/₄

Vì (x+¹/₂)²>=0 với mọi x

=>(x+¹/₂)²+³/₄>0 với mọi x

Vậy x²+x+1>0 với mọi x.

b)Ta có: -5-x²+2x

=-(x²-2x+5)

=-(x²-2x+1+4)

=-(x-1)²-4

Ta có:(x-1)²>=0 với mọi x

=>-(x-1)²<=0 với mọi x

=>-(x-1)²-4<0 với mọi x

Vậy -5-x²+2x<0 với mọi x

a) x²+x+1 =  \(x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

= \(x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}\) 

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

Do \(\left(x+\frac{1}{2}\right)^2\le0\)vs mọi x => \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)vs mọi x

=> x^2 + x + 1 > 0 vs mọi x

b) -5-x^2 + 2x = -(x^2 - 2x + 5) = \(-\left(x^2-2x+1+4\right)=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\)

Do \(-\left(x-1\right)^2\le0\)vs mọi x=> \(-\left(x-1\right)^2-4< 0\)vs mọi x 

=> -5-x^2+2x<0 vs 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 !!

a ) \(4x^2+2x+1=\left(2x\right)^2+2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

b ) \(x^2+3x+4=\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)

c ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)

Ta có : 4x² + 2x + 1

= (2x)² + 2.2x.\(\frac{1}{2}\) + \(\frac{1}{2}+\frac{3}{4}\)

= (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)

Mà : (2x + \(\frac{1}{2}\))² \(\ge0\forall x\)

=> (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\)

Hay : (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)  \(>0\forall x\)

Vậy 4x² + 2x + 1 \(>0\forall x\)