4x² - 8x + 5 >0 

(2x)² - 2. 2x.2 + 2² +1

(2x-2)²+1

Vì ( 2x-2) \(\ge\)0 mọi giá trị x => ( 2x-2)+1>0 với mọi giá trị x

Vậy 4x² - 8x + 5 > 0 với mọi giá trị của x

ta có 4x^2 - 8x + 5 = (2x)^2 - 2*2x *2  + 4 +1 = (2x - 2)^2 + 1 

do  (2x - 2)^2 >= 0 vs mọi x nên  (2x - 2)^2 + 1 > 0 với mọi x

(x-3)² +y² +1 >0

với mọi x;y

x² - 6x + 10 + y²

= x² - 6x + 9 + 1 + y²

= x² - 2.x.3 + 3² + 1 + y²

= (x - 3)² + 1 + y²

Ta có: (x - 3)² + 1 \(\ge\)1 và y² \(\ge\)0

=> (x - 3)² + 1 + y² \(\ge\)1 > 0 với mọi x, y (đpcm).

Giải:

a) \(x^2+xy+y^2+1\)

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

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

\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\ge1>0;\forall x\)

Vậy ...

HHuvg

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

Vì: \(\left(x-1\right)^2\ge0,\forall x\)

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

=>đpcm

b) \(x^2+7x+13=\left(x^2+7x+\frac{49}{4}\right)+\frac{3}{4}=\left(x+\frac{7}{2}\right)^2+\frac{3}{4}\)

Vì: \(\left(x+\frac{7}{2}\right)^2\ge0,\forall x\)

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

=>đpcm

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

Vì: \(-\left(x-\frac{1}{2}\right)^2\le0,\forall x\)

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

=>đpcm

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

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

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

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

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

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

giải giúp mik với