a) Đặt  \(A=x^2+4x+7\)

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

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

Mà  \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow A\ge3>0\)

b) Đặt  \(B=4x^2-4x+5\)

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

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

Mà  \(\left(2x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\)

c) Đặt  \(C=x^2+2y^2+2xy-2y+3\)

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

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

Mà  \(\left(x+y\right)^2\ge0\forall x;y\)

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

\(\Rightarrow C\ge2>0\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

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

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

\(=-\left(x^2+2x-8\right)\)

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

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

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

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

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

a) \(-9x^2+12x-15=-\left(9x^2-12x+4\right)-11=-\left(3x-2\right)^2-11\le11< 0\)

b) \(-2x^2+4x-9=-2\left(x^2-2x+1\right)-7=-2\left(x-1\right)^2-7\le-7< 0\)

c) \(xy-x^2-y^2-1=-\dfrac{1}{2}\left(2x^2+2y^2-2xy+2\right)=-\dfrac{1}{2}\left[\left(x-y\right)^2+x^2+y^2+2\right]< 0\)

a) x² - 8x + 19 = ( x² - 8x + 16 ) + 3 = ( x - 4 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) x² + y² - 4x + 2 = ( x² - 4x + 4 ) + y² - 2 = ( x - 2 )² + y² - 2 ≥ -2 ∀ x, y ( chưa cm được -- )

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

d) x² - 2xy + 2y² + 2y + 5 = ( x² - 2xy + y² ) + ( y² + 2y + 1 ) + 4 = ( x - y )² + ( y + 1 )² + 4 ≥ 4 > 0 ∀ x, y ( đpcm )

1/

\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)

\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)

2/

\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)

\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)

\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)

a) \(x^2 +x +1 = x^2 +x +1/4 +3/4 = (x+1/2)^2 +3/4\)

các câu khác dùng phương pháp tương tự

a) x^2 + x +1 = x^2 + x + 1/4 + 3/4 = ( x+ 1/2)^2 + 3/4

Vì (x+1/2)^2 >= 0 => (x+1/2)^2 + 3/4>=3/4 > 0

b) 4x^2 - 2x + 1 = (2x)^2 - 2x + 1/4 + 3/4 = (2x +1/2)^2 + 3/4

Vì (2x +1/2)^2 >=0 => (2x +1/2)^2 + 3/4 >= 3/4 > 0

c) x^4 -3x^2 + 9 = x^4 - 3x^2 + 9/4 + 25/4 = ( x^2+ 3/2)^2 + 9/4

Vì ( x^2+ 3/2)^2 >= 0 => ( x^2+ 3/2)^2 + 9/4 >=9/4 >0

d) x^2 + y^2 -2x-2y + 2xy +1

= ( x^2 + 2xy + y^2) - 2( x+y) +1

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

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

g) x^2+y^2+2(x-2y)+6

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

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

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

a. x² + x + 1 = x² + x +\(\frac{1}{4}+\frac{3}{4}\)= ( x +\(\frac{1}{2}\))² +\(\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

=> Đpcm

b. 4x² - 2x + 1 = 4x² - 2x +\(\frac{1}{4}+\frac{3}{4}\)= 4 ( x -\(\frac{1}{4}\))² +\(\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

=> Đpcm

c. x⁴ - 3x² + 9 = ( x² )² - 3x² + 3² = ( x² + 3x + 3 ) ( x² - 3x + 3 )

= [ ( x +\(\frac{3}{2}\))² +\(\frac{3}{4}\)] [ ( x -\(\frac{3}{2}\))² +\(\frac{3}{4}\)] > 0 với mọi x

=> Đpcm

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

= ( x + y )² - 2 ( x + y ) + 1²

= ( x + y - 1 )²\(\ge0\forall x;y\)

=> Đpcm

g. x² + y² + 2 ( x - 2y ) + 6 = x² + y² + 2x - 4y + 6

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

= ( x + 1 )² + ( y - 2 )² + 1\(\ge1>0\forall x;y\)

=> Đpcm