Toán Cho 1 ≤ a ≤ 2; 1 ≤ b ≤3; a + b + c = 11 Tìm Min P = a.b.c 10/07/2021 By Amara Cho 1 ≤ a ≤ 2; 1 ≤ b ≤3; a + b + c = 11 Tìm Min P = a.b.c
`abc=ab(11−a−b)=11ab−ab(a+b)` Ta có `(a−1)(b−1)≥0` `⇒a+b≤ab+1` `abc≥11ab−(ab+1)ab=−(ab)^2+10ab` `P=−(ab)^2+10ab−9+9=(ab−1)(9−ab)+9≥9` Dấu “=” xảy ra khi `a=b=1,c=9` #quangthinh347 Trả lời
`abc=ab(11−a−b)=11ab−ab(a+b)`
Ta có `(a−1)(b−1)≥0`
`⇒a+b≤ab+1`
`abc≥11ab−(ab+1)ab=−(ab)^2+10ab`
`P=−(ab)^2+10ab−9+9=(ab−1)(9−ab)+9≥9`
Dấu “=” xảy ra khi `a=b=1,c=9`
#quangthinh347