a Ta có 4x2 - 4x + 3 = (4x2 - 4x + 1) + 2 = (2x - 1)2 + 2 \(\ge\)2 > 0 (đpcm)
b) Ta có y - y2 - 1
= -(y2 - y + 1)
= -(y2 - y + 1/4) - 3/4
= -(y - 1/2)2 - 3/4 \(\le-\frac{3}{4}< 0\)(đpcm)
a) 4x2 - 4x + 3 = ( 4x2 - 4x + 1 ) + 2 = ( 2x - 1 )2 + 2 ≥ 2 > 0 ∀ x ( đpcm )
b) y - y2 - 1 = -( y2 - y + 1/4 ) - 3/4 = -( y - 1/2 ) - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )
4x2 - 4x + 3
= 4x2 - 4x + 1 + 2
= 4 ( x -\(\frac{1}{2}\))2 + 2\(\ge2\)
=> Đpcm
y - y2 - 1
= - y2 + y -\(\frac{1}{4}-\frac{3}{4}\)
= - ( y -\(\frac{1}{2}\))2 -\(\frac{3}{4}\le-\frac{3}{4}\)
=> Đpcm
\(4x^2-4x+3=4x^2-4x+1+2\)
\(=4\left(x-\frac{1}{2}\right)^2+2\ge2\left(đpcm\right)\)
\(b)y-y^2-1\)
\(=-y^2+y-\frac{1}{4}-\frac{3}{4}\)
\(=-\left(y-\frac{1}{2}\right)^2-\frac{3}{4}\le\frac{-3}{4}< 0\left(đpcm\right)\)