a) \(x^2-4x+5\)

= \(\left(x^2-2.2x+4\right)+1\)

= \(\left(x-2\right)^2+1\)

Ta co: \(\left(x-2\right)^2>=0\)

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

b) \(x^2-4xy+5y^2\)

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

= \(\left(x-2y\right)^2+y^2\)

Ta co: \(\left(x-2y\right)^2>=0\)

            \(y^2>=0\)

=> \(\left(x-2y\right)^2+y^2>=0\)

c) \(3-2x-x^2\)

= \(-\left(x^2+2x\right)+3\)

= \(-\left(x^2+2.1x+1-1\right)+3\)

= \(-\left(x+1\right)^2+4\)

= 

Hình như câu này sai đề ...

a) \(x^2-4x+5\)

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

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

b) \(x^2-4xy+5y^2\)

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

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

Dấu = xảy ra khi: \(x=y=0\)

c) \(-3-2x-x^2\)

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

\(=-2-\left(x+1\right)^2=-\left[2+\left(x+1\right)^2\right]< 0\)

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

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

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

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

https://olm.vn/hoi-dap/detail/88061957704.html bạn tham khảo câu hỏi này 

a) \(x^2+5y^2+2x-4xy-10y+14\)

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

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

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

Vì \(\left(x-2y+1\right)^2\ge0\)

      \(\left(y-3\right)^2\ge0\)

 \(\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+4\ge4>0\)với mọi x,y (ĐPCM)
b) \(5x^2+10y^2-6xy-4x-2y+3\)

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

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

Vì \(\left(2x-1\right)^2\ge0\)

      \(\left(x-3y\right)^2\ge0\)

       \(\left(y-1\right)^2\ge0\)

 \(\Rightarrow\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-1\right)^2+1\ge1>0\)vợi mọi x,y (ĐPCM)

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.

\(VT=x^2+2x\left(1-2y\right)+\left(1-2y\right)^2+\left(5y^2-\left(1-2y\right)^2-10y+14\right)\)

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

Ta có

x^2+5y^2+2x-4xy-10y+14 
=[x^2+2x(1-2y)+(1-2y)^2]+y^2-6y+13 
=(x+1-2y)^2+(y^2-2y.3+9)+4 
=(x+1-2y)^2+(y-3)^2+4. 
mà

(x+1-2y)^2 > hoặc=0 với mọi x,y thuộc R 
và (y-3)^2 > hoặc=0 với mọi y thuộc R 
=> (x+1-2y)^2+(y-3)^2+4 > hoặc =4 với mọi x,y thuộc R 
=> (x+1-2y)^2+(y-3)^2+4 >0 với mọi x,y thuộc R.

Akai Shuichi:chép sai đề rồi con ơi :D

\(a,x^2+5y^2+2x-4xy-10y+14\)

\(=x^2+2x-4xy+5y^2-10y+14\)

\(=x^2+2x\left(1-2y\right)+5y^2-10y+14\)

\(=x^2+2.x.\left(1-2y\right)+\left(1-2y\right)^2+5y^2-10y-\left(1-2y\right)^2+14\)

\(=\left(x+1-2y\right)^2+5y^2-10y-\left(1-4y+4y^2\right)+14\)

\(=\left(x+1-2y\right)^2+5y^2-10y-1+4y-4y^2+14\)

\(=\left(x+1-2y\right)^2+y^2-6y+13=\left(x+1-2y\right)^2+y^2-2.y.3+9+4\)

\(=\left(x+1-2y\right)^2+\left(y-3\right)^2+4\ge4>0\) với mọi x,y (đpcm)

b,tương tự