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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm GTLN: M=-x²+6x+8 N=7-4x-5x² Tìm GTNN M=x²+5x+6 N=7-4x+5x²

Toán Lớp 8: Tìm GTLN:
M=-x²+6x+8
N=7-4x-5x²
Tìm GTNN
M=x²+5x+6
N=7-4x+5x²

Comments ( 1 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    M = -x^2 + 6x + 8
    = -(x^2-6x-8)
    = -(x^2-6x+9-17)
    = -(x^2-2*x*3+3^2) + 17
    = -(x-3)^2 + 17
    Vì -(x-3)^2 \le 0 AAx
    => -(x-3)^2 + 17 \le 17 AAx
    Dấu = xảy ra khi :
    (x-3)^2 = 0
    => x – 3 = 0
    => x = 3
    Vậy M_\text{max} = 17 khi x = 3
    N = 7 – 4x – 5x^2
    = -5x^2 – 4x + 7
    = -(5x^2+4x-7)
    = -(x^2+4/5x+4/25-39/25)
    = -(x^2+4/5x+4/25) + 39/25
    = -[x^2+2*x*2/5+(2/5)^2] + 39/25
    = -(x+2/5)^2 + 39/25
    Vì -(x+2/5)^2 \le 0 AAx
    => -(x+2/5)^2 + 39/25 \le 39/25 AAx
    Dấu = xảy ra khi :
    (x+2/5)^2 = 0
    => x + 2/5 = 0
    => x = -2/5
    Vậy N_\text{max} = 39/25 khi x = -2/5
    M = x^2 + 5x + 6
    = x^2 + 5x + 25/4 – 1/4
    = [x^2+2*x*5/2+(5/2)^2] – 1/4
    = (x+5/2)^2 – 1/4
    Vì (x+5/2)^2 \ge 0 AAx
    => (x+5/2)^2 – 1/4 \ge -1/4 AAx
    Dấu = xảy ra khi :
    (x+5/2)^2 = 0
    => x + 5/2 = 0
    => x = -5/2
    Vậy M_\text{min} = -1/4 tại x = -5/2
    N = 7 – 4x + 5x^2
    = 5x^2 – 4x + 7
    = x^2 – 4/5x + 7/5
    = x^2 – 4/5x + 4/25 + 31/25
    = [x^2-2*x*2/5+(2/5)^2] + 31/25
    = (x-2/5)^2 + 31/25
    Vì (x-2/5)^2 \ge 0 AAx
    => (x-2/5)^2 + 31/25 \ge 31/25 AAx
    Dấu = xảy ra khi :
    (x-2/5)^2 = 0
    => x – 2/5 = 0
    => x = 2/5
    Vậy N_\text{min} = 31/25 tại x = 2/5

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222-9+11+12:2*14+14 = ? ( )