\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2x+1\right)+8\)

\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+8>0\forall x;y\)  (do \(\left(x-3y\right)^2\ge0;\left(2x-1\right)^2\ge0;\left(y-1\right)^2\ge0\forall x;y\)

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

c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)

a) \(A=x^2+2x+3=x^2+2x+1+2\)

\(=\left(x+1\right)^2+2\ge2\)

Vậy A luôn dương với mọi x

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

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

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

Vậy B luôn âm với mọi x

a)\(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)

Vậy x² +2x+3 luôn dương.

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

Vậy -x² +4x-5 luôn luôn âm.

a.x²+ 2x+ 3

=x²+ 2.x.1+ 1²- 1²+ 3

= (x+1)² -1+3

= (x+1)²+ 2

Ta có: (x+1)² ≥0

           (x+1)²+ 3≥ 3>0

⇒x²+ 2x+ 3>0 mọi x

Vậy x²+ 2x+3>0 mọi x

b. -x²+ 4x- 5

= - (x²- 4x +5)

= - (x²- 2.x.2+ 2²- 2²+ 5)

= - ((x- 2)²- 4+ 5)

= - ((x- 2)²+1)

= -(x- 2)² -1

Ta có: (x-2)² ≥0

         - (x-2)² ≤0

         - (x-2)² +1≤ 1

⇒ -x²+ 4x- 5 <0 mọi x

Vậy -x²+ 4x- 5 <0 mọi x

Ta có A = -x² + 4x - 6 - y² - 2y 

= -(x² - 4x + 4) - (y² + 2y + 1) - 1

= -(x - 2)² - (y + 1)² - 1 \(\le-1< 0\)

=> A < 0 với mọi x ; y

A = -x² + 4x - 6 - y² - 2y 

= -( x² - 4x + 4 ) - ( y² + 2y + 1 ) - 1

= -( x - 2 )² - ( y - 1 )² - 1 ≤ -1 < 0 ∀ x, y

=> đpcm

\(A=-x^2+4x-6-y^2-2y\)

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

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

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

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

mà \(\left(x-2\right)^2\ge0\forall x\)

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

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

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\)

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

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

\(\Rightarrow A< 0\)

Vậy A luôn có giá trị âm với mọi x,y

a Ta có 4x² - 4x + 3 = (4x² - 4x + 1) + 2 = (2x - 1)² + 2 \(\ge\)2 > 0 (đpcm)

b) Ta có y - y² - 1 

= -(y² - y + 1)

= -(y² - y + 1/4) - 3/4

= -(y - 1/2)² - 3/4 \(\le-\frac{3}{4}< 0\)(đpcm)

a) 4x² - 4x + 3 = ( 4x² - 4x + 1 ) + 2 = ( 2x - 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

b) y - y² - 1 = -( y² - y + 1/4 ) - 3/4 = -( y - 1/2 ) - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )

4x² - 4x + 3 

= 4x² - 4x + 1 + 2

= 4 ( x -\(\frac{1}{2}\))² + 2\(\ge2\)

=> Đpcm

y - y² - 1

= - y² + y -\(\frac{1}{4}-\frac{3}{4}\)

= - ( y -\(\frac{1}{2}\))² -\(\frac{3}{4}\le-\frac{3}{4}\)

=> Đpcm 

\(A=x^2+10y^2+2xy-6y+5\)

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

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

Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(3y+1\right)^2\ge0\\4>0\end{cases}}\)

=> A luôn dương với mọi x ; y

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

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

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

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

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

Mà \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\frac{3}{4}< 0\end{cases}}\)

=> B luôn âm với mọi x

a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)