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

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

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

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

\(A=x^2+y^2-4x+2y+7\)

\(=x^2+y^2-4x+2y+4+1+3\)

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

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

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+3\ge3>0\forall x,y\)

A= x²+y²-4x+2y+7

= (x²-4x+4)+(y²+2y+1)+2

= (x-2)²+(y+1)²+2

Ta thấy: (x-2)²\(\ge0\)

(y+1)²\(\ge0\)

\(\Rightarrow\)(x-2)²+(y+1)²+2\(\ge2\)

\(\Rightarrow\)A\(\ge2\)

Vậy A>0 \(\forall x,y\)

\(A=x^2+y^2-4x+2y+7\)

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

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

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

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+2\ge2>0\forall x,y\)

a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)

\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)

\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)

nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)

b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

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

Vì \(\left(x-y\right)^2\ge0\forall x;y\)

\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)

nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)

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) \(-\left(x^2-6x+10\right)=-\left(x^2-6x+9+1\right)=-\left[\left(x-3\right)^2+1\right]\le-1< 0\forall x\)

BĐT đúng

b) \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

BĐT đúng

c)Dấu "=" ko xảy ra???

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

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

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

a. −x² + 6x - 10

= x² − 6x) − 10

= x² − 2.x.3 + 3² − 9) − 10

= x − 3)² + 9 − 10

= x − 3)² −1

Vì x − 3)² ≥ 0 ∀ x ⇒ x − 3)² ≤ 0 ⇒ x − 3)² −1 ≤ −1

Vậy x − 3)² −1 < 0 ⇒ −x² + 6x - 10 luôn âm với mọi x

b. x² + x + 1

= x² + 2.x.\(\frac{1}{2}\)+ (\(\frac{1}{2}\))²

)² + \(\frac{3}{4}\)

Vì (x + \(\frac{1}{2}\))² ≥ 0 ∀ x ⇒ (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\) ∀ x

Vậy (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ 0 hay x² + x + 1 > 0 ∀ x.

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) 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

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

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

A=,x²-4xy-4y² +3

= (x-2y)²+3

do ( x-2y)²\(\ge0\forall x;y\)

=> (x-2y)²+3\(\ge3\)

=> A\(\ge3\)

vậy A \(>0\) với mọi số thực x;y

Bổ sung câu trả lời của bạn kuroba kaito

Khi và chỉ khi x - 2y =0

x =2y

Ta có: x² - 4xy + 4y² +3 > 0 với \(\forall\) x,y

\(\Leftrightarrow\) (x² - 4xy + 4y²) + 3 > 0 với \(\forall\) x,y

\(\Leftrightarrow\) (x - y)² +3 > 0 với \(\forall\) x,y

Ta thấy: (x - y)² \(\ge\) 0 với \(\forall\) x,y

3 > 0

\(\Rightarrow\) (x - y)² + 3 > 0 với \(\forall\) x,y

\(\Rightarrow\) x² - 4xy + 4x² + 3 > 0 với \(\forall\) x,y