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

Toán Lớp 8: tìm gtnn của (x+1)(x+2)(x+3)(x+4)

Toán Lớp 8: tìm gtnn của (x+1)(x+2)(x+3)(x+4)

Comments ( 2 )

  1. (x+1)(x+2)(x+3)(x+4)
    =(x^2+5x+4)(x^2+5x+6)
    Đặt x^2+5x+5=t
    =(t-1)(t+1)
    =t^2-1>=-1AAx in RR
    =>min=-1<=>x^2+5x+5=0<=>x=(-5+-sqrt5)/2

  2. #tnvt
    (x+1)(x+2)(x+3)(x+4)
    =(x+1)(x+4)(x+2)(x+3)
    =(x^2+5x+4)(x^2+5x+6)
    Đặt x^2+5x+5=a
    =(a-1)(a+1)
    =a^2-1
    Vì a^2>=0<=>a^2-1>=-1
    Dấu = xảy ra khi a=0<=>x^2+5x+5=0
    <=>x^2+2.x. 5/2+25/4-5/4=0
    <=>(x+5/2)^2-5/4=0
    <=>(x+5/2-\frac{\sqrt{5}}{2})(x+5/2+\frac{\sqrt{5}}{2})=0
    <=>[(x=\frac{\sqrt{5}-5}{2}),(x=\frac{-\sqrt{5}-5}{2}):}
    Vậy GTNNNN=-1 khi x=\frac{+-\sqrt{5}-5}{2}

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