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

Toán Lớp 6: a) P= 2 mũ 100 – 2 mũ 99 – 2 mũ 98 – – – 2 mũ 3 – 2 mũ 2 – 2 b) P= 2.4.8.16.32.64.128.256.512.1024:2 mũ 52

Home/ Toán Lớp 6: a) P= 2 mũ 100 – 2 mũ 99 – 2 mũ 98 – – – 2 mũ 3 – 2 mũ 2 – 2 b) P= 2.4.8.16.32.64.128.256.512.1024:2 mũ 52

Toán Lớp 6: a) P= 2 mũ 100 – 2 mũ 99 – 2 mũ 98 – – – 2 mũ 3 – 2 mũ 2 – 2 b) P= 2.4.8.16.32.64.128.256.512.1024:2 mũ 52

Tháng Bảy 27, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: a) P= 2 mũ 100 – 2 mũ 99 – 2 mũ 98 – …. – 2 mũ 3 – 2 mũ 2 – 2
b) P= 2.4.8.16.32.64.128.256.512.1024:2 mũ 52

Comments ( 2 )

  1. mytamhoang
    mytamhoang
    27 Tháng Bảy, 2022 at 5:27 chiều
    Reply
    Giải đáp:
    a)2                 b)8
    Lời giải và giải thích chi tiết: 
    a)   P=                  $2^{100}$ – $2^{99}$ – $2^{98}$ – ….. – $2^{3}$ – $2^{2}$ – 2   
         2P= $2^{101}$ – $2^{100}$ – $2^{99}$ – $2^{98}$ – ….. – $2^{3}$ – $2^{2}$
     2P-P= ( $2^{101}$ – $2^{100}$ $2^{99}$ – ….. – $2^{3}$ – $2^{2}$ ) – ( $2^{100}$ – $2^{99}$ – ….. – $2^{2}$ -2 )     
           P= $2^{101}$ – $2^{100}$ – $2^{99}$ – ….. – $2^{3}$ – $2^{2}$ – $2^{100}$ + $2^{99}$ + ….. + $2^{2}$ + 2 
           P= $2^{101}$ – $2^{100}$ – $2^{100}$ +2 
           P= $2^{101}$ – ( $2^{100}$ + $2^{100}$ ) +2 
           P= $2^{101}$ – 2× $2^{100}$ + 2 = $2^{101}$ – $2^{101}$ +2 = 2
    b)   P= 2×4×8×16×32×64×128×256×512×1024: $2^{52}$
          P= 2× $2^{2}$ × $2^{3}$ × ….. ×   $2^{8}$ × $2^{9}$ × $2^{10}$ : $2^{52}$
          P=(2× $2^{2}$ × $2^{3}$ × ….. ×   $2^{8}$ × $2^{9}$ × $2^{10}$ ) : $2^{52}$
          P= $2^{55}$ : $2^{52}$ = $2^{3}$ = 8

  2. kymmailien
    kymmailien
    27 Tháng Bảy, 2022 at 5:27 chiều
    Reply

    Đáp án + Giải thích các bước giải:

    ĐK:x≥0;x≠1ĐK:x≥0;x≠1

    a)A=x√x+1a)A=xx+1

    Thay x=16x=16  vào  AA

    ⇒A=16√16+1⇒A=1616+1

    =164+1=165=164+1=165

    b)B=3√x+1+11−√x+x+5x−1b)B=3x+1+11-x+x+5x-1

    =3√x+1−1√x−1+x+5(√x−1)(√x+1)=3x+1-1x-1+x+5(x-1)(x+1)

    =3(√x−1)−(√x+1)+(x+5)(√x−1)(√x+1)=3(x-1)-(x+1)+(x+5)(x-1)(x+1)

    =3√x−3−√x−1+x+5(√x−1)(√x+1)=3x-3-x-1+x+5(x-1)(x+1)

    =x+2√x+1(√x−1)(√x+1)=x+2x+1(x-1)(x+1)

    =(√x+1)2(√x−1)(√x+1)=(x+1)2(x-1)(x+1)

    =√x+1√x−1=x+1x-1

    c)K=A.Bc)K=A.B

    ⇒K=x√x+1.√x+1√x−1⇒K=xx+1.x+1x-1

    ⇒K=x√x−1⇒K=xx-1

    Để K≤4K≤4

    thì x√x−1≤4xx-1≤4

    ⇔x√x−1−4≤0⇔xx-1-4≤0

    ⇔x−4√x+4√x−1≤0⇔x-4x+4x-1≤0

    ⇔(√x−2)2√x−1≤0⇔(x-2)2x-1≤0

    Vì (√x−2)2≥0(x-2)2≥0

    ⇒√x−1

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

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