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

Toán Lớp 8: bài 3 chứng minh rằng a. 35^2005 – 35^2004 : 17 b. 43^2004 + 43^2005 : 11 c. 27^3 + 9^5 : 4

Toán Lớp 8: bài 3 chứng minh rằng
a. 35^2005 – 35^2004 : 17
b. 43^2004 + 43^2005 : 11
c. 27^3 + 9^5 : 4

Comments ( 2 )

  1. Giải đáp:
     a, Ta có:
    35^2005 – 35^2004 = 35^2004 (35 – 1)
                                 = 35^2004 .34
    Mà 34 $\vdots$ 17
    ⇒ 35^2004 .34 $\vdots$ 17
    Hay: 35^2005 – 35^2004 $\vdots$ 17 (đpcm)
    b, Ta có: 
    43^2004 + 43^2005 = 43^2004 (1 + 43)
                                  = 43^2004 .44
    Mà 44 $\vdots$ 11
    ⇒ 43^2004 .44 $\vdots$ 11
    Vậy 43^2004 + 43^2005 $\vdots$ 11 (đpcm)
    c, Ta có:
    27^3 + 9^5 = (3^3)^3 + (3^2)^5 = 3^9 + 3^10
                     = 3^9 (1 + 3) = 3^9 .4
    Mà 4 $\vdots$ 4
    ⇒ 3^9 .4 $\vdots$ 4
    Vậy 27^3 + 9^5 $\vdots$ 4 (đpcm)

  2. a,35^{2005}-35^{2004}=35^{2004} . (35-1)= 35^{2004} . 34
    Vì 34 chia hết 17 nên 35^{2004} . 34 \vdots 17 hay 35^{2005}-35^{2004} \vdots 17->đpcm
    b,43^{2004}+43^{2005}=43^{2004} . ( 1+43)= 43^{2004} . 44
    Vì 44\vdots 11 nên 43^{2004} . 44 \vdots 11 hay 43^{2004}+43^{2005} \vdots 11->đpcm
    c,27^3 + 9^5 = (3^3)^3 + (3^2)^5 = 3^{3.3}+3^{5.2}=3^9 + 3^10= 3^9 . ( 1+3)=3^9 .4
    Vì 4\vdots 4 nên 3^9 .4 \vdots 4 hay 27^3 + 9^5 \vdots 4->đpcm

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