`B = -5 + (x - 1/2)^2`
Ta có: `(x - 1/2)^2 ≥ 0 ∀x`
`=> (x - 1/2)^2 - 5 ≥ -5`
Dấu "=" xảy ra khi `(x - 1/2)^2 = 0`
`=> x - 1/2 = 0`
`=> x = 1/2`
Vậy MinB = -5 tại `x = 1/2`.
`E = |x| + |x - 1| + ... + |x - 99|`
`= (|x| + |x - 99|) + (|x - 1| + |x - 98|) + ... + (|x - 49| + |x - 50|)`
`= (|x| + |99 - x|) + (|x - 1| + |98 - x|) + ... + (|x - 49| + |50 - x|)`
`=> D ≥ |x + 99 - x| + |x - 1 + 98 - x| + ... + |x - 49 + 50 - x|`
`D ≥ 99 + 97 + ... + 1`
`D ≥ (1 + 99) . 50 : 2 = 2500`
Dấu "=" xảy ra khi `x (99 - x) ≥ 0 ; (x - 1) (98 - x) ≥ 0 ; ... ; (x - 49) (50 - x) ≥ 0`
`=> x (x - 99) ≤ 0 ; (x - 1) (x - 98) ≤ 0 ; ... ; (x - 49) (x - 50) ≤ 0`
`=> 0 ≤ x ≤ 99 ; 1 ≤ x ≤ 98 ; ... ; 49 ≤ x ≤ 50`
`=> 49 ≤ x ≤ 50`
Vậy MinE = 2500 tại `49 ≤ x ≤ 50`.
