`A=x^2 -4x+18`

`=x^2 -4x+4+14`

`=(x-2)^2 +14`

Có `(x-2)^2 >=0 AAx`

`=> (x-2)^2 +14>= 14>0 AAx`

Vậy ....

`B=x^2 -x+2`

`=x^2 -x+1/4+7/4`

`=(x-1/2)^2 +7/4`

có `(x-1/2)^2 >=0 AAx`

`=> (x-1/2)^2 +7/4>=7/4>0 AAx`

Vậy ...........

`C=x^2 +2y^2 -2xy-2y+15`

`=x^2 -2xy+y^2 +y^2 -2y+1+14`

`=(x-y)^2 +(y-1)^2 +14`

Có `(x-y)^2 >=0 AAx,y` ;   `(y-1)^2 >=0 AAy`

`=>(x-y)^2 +(y-1)^2 +14 >=14>0 AAx;y`

Vậy

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

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

                                  \(=-1-\left(\frac{1}{2}x-1\right)^2\le-1\left(đpcm\right)\)

b)\(-3x^2-6x-9=-3\left(x^2-2x+1+2\right)\)

                                  \(=-6-3\left(x-1\right)^2\le-6\left(đpcm\right)\)

c)\(-2x^2+3x-6=-2\left(x^2-\frac{3}{2}x+3\right)\)

                                  \(=-2\left(x^2-2.\frac{3}{4}x+\frac{9}{16}+\frac{39}{16}\right)\)

                                     \(=-\frac{39}{8}-2\left(x-\frac{3}{4}\right)^2\le-\frac{39}{8}\)

d) tương tự

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

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Leftrightarrow-\left(x-\frac{1}{2}\right)^2\le0\Leftrightarrow-\left(x-\frac{1}{2}\right)^2-1\le-1< 0\)

=> biểu thức luôn âm

các câu sau tương tự, nếu bạn chưa rõ thì có thể hỏi lại mình

A=x²-6x+10

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

\(\Rightarrow A\) luôn dương

A = x² - 6x + 10

= ( x² - 6x + 9 ) + 1 

= ( x - 3 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

B = x² + x + 5

= ( x² + x + 1/4 ) + 19/4

= ( x + 1/2 )² + 19/4 ≥ 19/4 > 0 ∀ x ( đpcm )

C = 4x² + 4x + 2 

= 4( x² + x + 1/4 ) + 1

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

D = ( x - 3 )( x - 5 ) + 4

= x² - 8x + 15 + 4

= ( x² - 8x + 16 ) + 3 

= ( x - 4 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

E = x² - 2xy + 1 + y²

= ( x² - 2xy + y² ) + 1 

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

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)

a) \(x^2+6x+10\)

\(=\left(x^2+2.3x+9\right)+1\)

\(=\left(x+3\right)^2+1\ge1>0\)

\(\Rightarrow DPCM\)

b) \(x^2-x+1\)

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

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

\(\Rightarrow DPCM\)

c) \(x^4-4x^2+5\)

\(=\left[\left(x^2\right)^2-2.2.x^2+2^2\right]+1\)

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

\(\Rightarrow DPCM\)

\(a,A=4x^2-20x+27=\left(2x\right)^2-2.2x.5+5^2+2\)\(=\left(2x-5\right)^2+2\)

Mà \(\left(2x-5\right)^2\ge0\Rightarrow\left(2x-5\right)^2+2>0\Rightarrow A>0\)

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

Mà \(\left(x-\frac{1}{4}\right)^2\ge0\Rightarrow\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\Rightarrow B>0\)

\(c,C=x^2+4x+y^2-6y+15=x^2+4x+4+y^2-6y+9+2\)

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

Mà \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+2>0\Rightarrow C>0\)

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

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

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