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

Toán Lớp 9: `\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}` tìm gtnn

Toán Lớp 9: \sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}
tìm gtnn

Comments ( 2 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
    \sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}(x≥1)
    =\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}
    =\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}
    =|\sqrt{x-1}-1|+|\sqrt{x-1}+1|
    =|1-\sqrt{x-1}|+|\sqrt{x-1}+1|
    |1-\sqrt{x-1}|+|\sqrt{x-1}+1|≥|1-\sqrt{x-1}+\sqrt{x-1}+1|
    ⇔|1-\sqrt{x-1}|+|\sqrt{x-1}+1|≥2
    Dấu “=” xảy ra khi
    (1-\sqrt{x-1})(\sqrt{x-1}+1)≥0
    \sqrt{x-1}+1≥1>0∀xtmdk
    ⇒1-\sqrt{x-1}≥0
    ⇔\sqrt{x-1}≤1
    ⇔x-1≤1
    ⇔x≤2
    Kết hợp điều kiện xác định
    Vậy $gtnn$ biểu thức là 2 khi 1≤x≤2
     

  2. Giải đáp :
    A=sqrt(x-2sqrt(x-1))+sqrt(x+2sqrt(x-1))
    A=sqrt((x-1)-2sqrt(x-1)+1)+sqrt((x-1)+2sqrt(x-1)+1)
    A=sqrt((sqrt(x-1)-1)^2)+sqrt((sqrt(x-1)+1)^2)
    A=|sqrt(x-1)-1|+|sqrt(x-1)+1|
    A=|1-sqrt(x-1)|+|sqrt(x-1)+1|
    A>=|1-sqrt(x-1)+sqrt(x-1)+1|
    A>=|1+1|
    A>=|2|
    A>=2
    Xảy ra dấu “=” :
    (sqrt(x-1)-1)(sqrt(x-1)+1)=0
    <=>x-1-1=0
    <=>x-2=0
    <=>x=2
    Vậy : A_min=2 khi x=2

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

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