tìm GTNN
B= -5 + ( x- 1/2 )^2
E= |x| + |x-1| + |x-2| ….|x-98| + |x-99|
tìm GTNN B= -5 + ( x- 1/2 )^2 E= |x| + |x-1| + |x-2| ….|x-98| + |x-99|
By Katherine
By Katherine
tìm GTNN
B= -5 + ( x- 1/2 )^2
E= |x| + |x-1| + |x-2| ….|x-98| + |x-99|
Đáp án:
`B=-5+(x-1/2)^2`
Vì `(x-1/2)^2>=0`
`=>(x-1/2)^2-5>=-5`
Hay `B>=-5`
Dấu “=” xảy ra khi `x-1/2=0<=>x=1/2`.
Vậy `min_B=-5<=>x=1/2`
`E=|x|+|x-1|+|x-2|+….|x-98|+|x-99|`
`=(|x|+|99-x|)+(|x-1|+|98-x|)+…..+(|x-49|+|50-x|)`
Áp dụng bđt `|A|+|B|>=|A+B|`
`=>|x|+|99-x|>=|x+99-x|=99`
`|x-1|+|98-x|>=|x-1+98-x|=97`
`………………………………………………`
`|x-49|+|50-x|>=1`
`=>E>=99+97+……………..+1`
`=>E>=((99+1).99)/2=4950`.
Dấu “=” xảy ra khi:\(\begin{cases}x(99-x) \ge 0\\(x-1)(98-x) \ge 0\\(x-49)(50-x) \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x(x-99) \le 0\\(x-1)(x-98) \le 0\\…………….\\(x-49)(x-50) \le 0\\\end{cases}\)
`<=>` \(\begin{cases}0 \le x \le 99\\1 \le x \le 98\\…………..\\49 \le x \le 50\\\end{cases}\)
`<=>49<=x<=50`.
`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`.