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