a. x^2 luôn lớn hơn hoặc bằng x.

x^2 lớn hơn hoặc bằng 0 

x^2-x cũng sẽ lớn hơn hoặc bằng 0

Khi cộng thêm 1 nó lớn hơn hoặc pằng 1=> dương

x^2-8x+20=(x^2-8x+16)+4

                 =(x-4)^2+4>0(vì (x-4)^2>=0)

4x^2-12x+11=4x^2-12x+9+2

                     =(2x-3)^2+2>0

x^2-x+1=x^2-x+1/4+3/4

             =(x-1/2)^2+3/4>0

x^2-2x+y^2+4y+6

=x^2-2x+1+y^2+4y+4+1

=(x-1)^2+(y+2)^2+1>0

a: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(4x^2-12x+11\)

\(=4x^2-12x+9+2\)

\(=\left(2x-3\right)^2+2>0\forall x\)

c: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

d: Ta có: \(x^2-2x+y^2+4y+6\)

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

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

Ta có : x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1

Mà :  (x + 1)² \(\ge0\forall x\)

Nên : (x + 1)² + 1 \(\ge1\forall x\)

Vậy  x² + 2x + 2 luôn dương 

\(A=x^2+2x+2\)

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

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

Nhận thấy   \(\left(x+1\right)^2\ge0\)

=>  \(\left(x+1\right)^2+1>0\)

=> A luôn dương

a) \(A=x^2+2x+2\)

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

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

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

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

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

\(=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4>0\forall x,y\)

\(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

\(C=x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(D=x^2-2x+y^2+4y+6=x^2-2x+1+y^2+4y+4+1\)

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

\(E=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

\(A=2x^2-20x+7=2\left(x^2-10x+25\right)-43=2\left(x-5\right)^2-43\ge-43\left(\forall x\right)\)

=> Chưa thể khẳng định A dương

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

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

\(B=\left(3x-y\right)^2+y^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

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

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

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

=> đpcm

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

=> đpcm

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

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

Xét nữa là xong

\(a.\)

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

\(A=\left(3x\right)^2-2\cdot3x\cdot y+y^2+y^2+1\)

\(A=\left(3x-y\right)^2+\left(y^2+1\right)\ge0\)

\(b.\)

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

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

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

\(c.\)

\(C=x^2-2x+2\)

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

\(C=\left(x-1\right)^2+1\ge1\)

a) A=9x²-6xy+2y²+1

    A=(3x)²-2.3x.y+y²+y²+1

    A=(3x-y)²+(y²+1)≥0

Câu b, c tương tự câu a

a) Ta có: \(A=9x^2-6xy+2y^2+1\)

\(=\left(3x-1\right)^2+2y^2\ge0\forall x,y\)

b) Ta có: \(B=x^2-2x+y^2+4y+6\)

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

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

c) Ta có: \(C=x^2-2x+2\)

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

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

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

b) 4x² - 2x + 1 = 4( x² - 1/2x + 1/16 ) + 3/4 = 4( x - 1/4 )² + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )

c) x⁴ - 3x² + 9 (*)

Đặt t = x²

(*) <=> t² - 3t + 9 = ( t² - 3t + 9/4 ) + 27/4 = ( t - 3/2 )² + 27/4 = ( x² - 3/2 )² + 27/4 ≥ 27/4 > 0 ∀ x ( đpcm )

d) x² + y² - 2x - 4y + 6 = ( x² - 2x + 1 ) + ( y² - 4y + 4 ) + 1 = ( x - 1 )² + ( y - 2 )² + 1 ≥ 1 > 0 ∀ x, y ( đpcm )

e) x² + y² - 2x - 2y + 2xy + 2 = ( x² + 2xy + y² - 2x - 2y + 1 ) + 1

                                              = [ ( x² + 2xy + y² ) - ( 2x + 2y ) + 1 ] + 1 

                                              = [ ( x + y )² - 2( x + y ) + 1² ] + 1

                                              = ( x + y - 1 )² + 1 ≥ 1 > 0 ∀ x, y ( đpcm )

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

b) \(4x^2-2x+1=4\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{3}{4}=4\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)

c) \(x^4-3x^2+9=\left(x^4-3x^2+\frac{9}{4}\right)+\frac{27}{4}=\left(x^2-\frac{3}{2}\right)^2+\frac{27}{4}>0\left(\forall x\right)\)

d) \(x^2+y^2-2x-4y+6\)

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

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

e) \(x^2+y^2-2x-2y+2xy+2\)

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

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

A = x(x - 6) + 10

A = x^2 - 6x + 9 + 1

A = (x - 3)^2 + 1 > 1

B = x^2 - 2x + 9y^2 - 6y + 3

B = (x^2 - 2x + 1) + (9y^2 - 6y + 1) + 1

B = (x - 1)^2 + (3y - 1)^2 + 1 > 1

a) 4x² - 12x + 11=4x²-12x+9+2=(2x-3)²+2

vì (2x-3)²\(\ge\)0

nên (2x-3)²+2 dương với mọi x

=>4x² - 12x + 11luôn luôn dương với mọi x

b) x² - 2x + y² + 4y + 6

=x²-2x+1+y²+4y+4+1

=(x-1)²+(y+2)²+1

vì (x-1)²\(\ge\)0 ; (y+2)²\(\ge\)0

nên (x-1)²+(y+2)²+1 dương với mọi x;y

=>x² - 2x + y² + 4y + 6  luôn dương với mọi x;y

a/B=x²+2x+2013

a/ chứng minh x²+2x+2 luôn dương với mọi x

b/ chứng minh x²-2x+y²+4y+6 luôn dương với mọi x;y

c/ chứng minh -x²-8x-15 luôn âm với mọi x

d/ chứng minh 8-x²-y²-4x+10 luôn âm với mọi x;y