Cho 1 ≤ a ≤ 2; 1 ≤ b ≤3; a + b + c = 11 Tìm Min P = a.b.c

Cho 1 ≤ a ≤ 2; 1 ≤ b ≤3; a + b + c = 11
Tìm Min P = a.b.c

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  1. `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

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