a) `x^2 + 2x + 9`
`= x^2 + 2x.1 + 1^{2} + 8`
`= ( x + 1 )^2 + 8 \ge 8`
Dấu `=` xảy ra `<=> x= -1`
Vậy...
b) `4x^2 + 4x + 10`
`= [ (2x)^2 + 2.2x.1 + 1^2 ] + 9`
`= ( 2x + 1 )^2 + 9 \ge 9`
Dấu `=` xảy ra khi `x = - \frac{1}{2}`
Vậy...
c) `9x^2 + 6x + 17`
`= [ (3x)^2 + 2.3x.1 + 1^2 ] + 16`
`= (3x + 1 )^2 + 16 \ge 16`
Dấu `=` xảy ra `<=> x = - \frac{1}{3}`
Vậy...
d) `16x^2 + 8x + 101`
`= [ ( 4x)^2 + 2.4x.1 + 1^2 ] + 100`
`= ( 4x + 1 )^2 + 100 \ge 100`
Dấu `=` xảy ra `<=> x = - \frac{1}{4}`
Vậy...
