Bài 3
a) A = - 9x^2 - 6x - 9
        = -  ( 9x^2 + 6x + 1 ) - 8
        = - ( 3x + 1 )^2 - 8 < 0
b) B = - 1 - x^2 + x
        = - ( x^2 - x + 1/4 ) - 3/4
        = - ( x - 1/2 )^2 - 3/4 < 0
c) C = - 2x^2 - 2x - 1
        = - 2.( x^2 + x + 1/4 ) - 1/2
        = - ( x + 1/2 )^2 - 1/2 < 0
e) E = -4x^2 - 4x - 2019
       = - ( 4x^2 + 4x + 1 ) - 2018
       = - ( 2x + 1 )^2 - 2018 < 0
 

2
e) E >0
<=> 4x^2 - 4x +2019 >0
<=> 4x^2 - 4x + 1 +2018 >0
<=> (4x^2 - 4x + 1) >-2018
<=> (2x - 1)^2 > -2018 với mọi x
f) F > 0
<=> x^2 + 5x + 7 > 0
<=> x^2 + 5x + 25/4 + 3/4 > 0
<=> (x^2 + 5x + 25/4) > -3/4
<=> (x + 5/2)^2 > -3/4 với mọi x
g) G > 0
<=> -x^4 + 4x^2 - 104 > 0
<=> -(x^4 - 4x^2 + 4) - 100 > 0
<=> (x^4 - 4x^2 + 4) < -100
<=> (x^2 - 2)^2 < -100

Bài 4
a) A = 9x^2 - 6x + 2  
        = 9x^2 - 6x + 1 + 1
        = ( 3x + 1 )^2 + 1 ≥ 1
Vậy GTNN của A = 1 <=> x = -1/3
b) B = x^2 + x + 1
        = x^2 + x + 1/4 + 3/4
        = ( x + 1/2 )^2 + 3/4 ≥ 3/4
Vậy GTNN của B = 3/4 <=> x = -1/2
c) C = 2x^2 + 2x + 1
        = 2.( x^2 + x + 1/4 ) + 1/2
        = 2.( x + 1/2 )^2 + 1/2 ≥ 1/2
Vậy GTNN của C = 1/2 <=> x = -1/2
e) E = 4x^2 - 4x + 2019
       = 4x^2 - 4x + 1 + 2018
       = ( 2x - 1 )^2 + 2018 ≥ 2018
Vậy GTNN của E = 2018 <=> x = 1/2

Bài 1
d) x^2 + y^2 + 2x + 2y + 2.( x + 1 ).( y + 1 ) + 2
= ( x^2 + 2x + 1 ) + ( y^2 + 2y + 1 ) + 2.( x + 1 ).( y + 1 )
= ( x + 1 )^2 + 2.( x + 1 ).( y + 1 ) + ( y + 1)^2
= ( x + 1 + y + 1 )^2
= ( x + y + 2 )^2
f) x^2 + 2x.( y + 1 ) + y^2 + 2y + 1
= x^2 + 2x.( y + 1 ) + ( y + 1 )^2
= ( x + y + 1 )^2
e) x^2 - 2x.( y + 2 ) + y^2 + 4y + 4
= x^2 - 2x.( y + 2 ) + ( y + 2 )^2
= ( x + y + 2 )^2
b) ( x + 1 ).( x - 2 ).( x - 3 ).( x - 6 ) + 36
= [ ( x + 1 ).( x - 6 )].[ ( x - 2 ).( x - 3 )] + 36
= ( x^2 - 5x - 6 ).( x^2 - 5x + 6 ) + 36
= ( x^2 - 5x )^2 - 36 + 36
= ( x^2 - 5x )^2
 

Bài 1
a, (x+3)(x+4)(x+5)(x+6) + 1
= (x+3)(x+6)(x+4)(x+5) + 1
= (x^2 + 9x + 18)(x^2 + 9x + 20) + 1
Đặt x^2 + 9x + 19 = y ta có:
(y-1)(y+1) + 1
= y^2 - 1 + 1
= y^2
= (x^2 + 9x + 19)^2   (đpcm)

b, (x+1)(x-2)(x-3)(x-6) + 36
= (x+1)(x-6)(x-2)(x-3) + 36
= (x^2 - 5x - 6)(x^2 - 5x + 6) + 36
Đặt x^2 - 5x = y ta có:
(y-6)(y+6) + 36
= y^2 - 36 + 36
= y^2
= (x^2 - 5x)^2   (đpcm)

Bài 4
g, G = x^2 - 6x + 11
       = x^2 - 6x + 9 + 2
       = (x-3)^2 + 2 ≥ 2
Dấu "=" xảy ra <=> x-3 = 0
                      <=> x = 3
Vậy Min G = 2 <=> x = 3

h, H = x^2 - 3x + 5
       = x^2 - 3x + 9/4 + 5 - 9/4
       = (x-3/2)^2 + 11/4 ≥ 11/4
Dấu "=" xảy ra <=> x-3/2 = 0
                      <=> x = 3/2
Vậy Min H = 11/4 <=> x = 3/2

Bài 4
i, I = x^2 - 20x + 101
    = x^2 - 20x + 100 + 1
    = (x-10)^2 + 1 ≥ 1
Dấu "=" xảy ra <=> x-10 = 0
                      <=> x = 10
Vậy Min I = 1 <=> x = 10

l, L = x^4 - 4x^2 + 104
     = x^4 - 4x^2 + 4 + 100
     = (x^2 - 2)^2 + 100 ≥ 100
Dấu "=" xảy ra <=> x^2 - 2 = 0
                      <=> x^2 = 2
                      <=> x = ±√2
Vậy Min L = 100 <=> x = ±√2
 

m, M = (2x-1)^2 + (x+2)^2
        = 4x^2 - 4x + 1 + x^2 + 4x + 4
        = 5x^2 + 5 ≥ 5
Dấu "=" xảy ra <=> 5x^2 = 0
                      <=> x^2 = 0
                      <=> x = 0
Vậy Min M = 5 <=> x = 0

n, N = (x-2)^2 + 3(x+3)^2 + 5
       = x^2 - 4x + 4 + 3x^2 + 18x + 27 + 5
       = 4x^2 + 14x + 36
       = 4x^2 + 14x + 49/4 + 36 - 49/4
       = (2x+7/2)^2 + 95/4 ≥ 95/4
Dấu "=" xảy ra <=> 2x+7/2 = 0
                      <=> 2x = -7/2
                      <=> x = -7/4
Vậy Min N = 95/4 <=> x = -7/4