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

Toán Lớp 7: Bài 1 : Chứng minh rằng c) 1/1.2 + 1/2.3 + -+ 1/2007.2008 <1 d) A= 1+5+5^2+-+5^9/1+5+5^2+-.+5^8 Bài 2 : Tìm x a) 2008^x = 1

Home/ Toán Lớp 7: Bài 1 : Chứng minh rằng c) 1/1.2 + 1/2.3 + -+ 1/2007.2008 <1 d) A= 1+5+5^2+-+5^9/1+5+5^2+-.+5^8 Bài 2 : Tìm x a) 2008^x = 1

Toán Lớp 7: Bài 1 : Chứng minh rằng c) 1/1.2 + 1/2.3 + -+ 1/2007.2008 <1 d) A= 1+5+5^2+-+5^9/1+5+5^2+-.+5^8 Bài 2 : Tìm x a) 2008^x = 1

Tháng Mười 19, 2022 Toán học 1 Comment 0 views

Toán Lớp 7: Bài 1 : Chứng minh rằng
c) 1/1.2 + 1/2.3 + ….+ 1/2007.2008 <1 d) A= 1+5+5^2+....+5^9/1+5+5^2+.....+5^8 Bài 2 : Tìm x a) 2008^x = 1 b) 1/32^x . 16^x = 1024 c) 5^x + 5^x+2 = 650 d) 1/3 . 3^x + 5 . 1/3^x-1 e) 1/3^2 . 3^4 . 3^x =3^7 g) 9.27^x = 3^5 h) 1/2 . 2^x + 4.2^x = 9. 2^5

Comments ( 1 )

  1. maianh67
    maianh67
    19 Tháng Mười, 2022 at 10:59 chiều
    Reply
    $\\$
    Bài 1.
    c,
    1/(1.2)+1/(2.3)+…+1/(2007 . 2008)
    = 1 – 1/2+1/2-1/3+…+1/2007 – 1/2008
    = 1 – 1/2008
    Vì 1 – 1/2008 < 1
    -> 1/(1.2)+1/(2.3)+….+1/(2007 . 2008)<1
    d,
    Đặt B =1+5+…+5^9
    ->5B = 5 (1+5+…+5^9)
    ->5B = 5 +5^2+…+5^{10}
    ->5B-B=(5+5^2+…+5^{10})-(1+5+…+5^9)
    ->4B=5^{10}-1
    ->B=(5^{10}-1)/4
    Đặt C=1+5+…+5^8
    -> 5C=5(1+5+…+5^8)
    -> 5C=5+5^2+…+5^9
    ->5C-C=(5+5^2+…+5^9)-(1+5+…+5^8)
    -> 4C=5^9 – 1
    ->C=(5^9 – 1)/4
    A = (1+5+…+5^9)/(1+5+…+5^8)
    ->A = (5^{10}-1)/4 : (5^9 – 1)/4
    -> A = (5^{10}-1)/4 . 4/(5^9 – 1)
    ->A = (5^{10}-1)/(5^9 – 1)
    Vậy A = (5^{10}-1)/(5^9 – 1)
    $\\$
    Bài 2.
    a,
    2008^x = 1
    ->2008^x=2008^0
    ->x=0
    Vậy x=0
    b,
    1/32^x . 16^x = 1024
    -> (1/32)^x . (2^4)^x = 2^{10}
    -> (1/2)^{5x} . 2^{4x} = 2^{10}
    -> 2^{-5x} . 2^{4x}=2^{10}
    ->2^{-x}=2^10
    ->x=-10
    Vậy x=-10
    c,
    5^x + 5^{x+2}=650
    ->5^x + 5^x . 5^2 =650
    ->5^x . (1+5^2)=650
    ->5^x . 26=650
    ->5^x=25
    ->5^x = 5^2
    ->x=2
    Vậy x=2
    d,
    Xem lại đề.
    e,
    1/3^2 . 3^4 . 3^x =3^7
    -> 3^2 . 3^x = 3^7
    ->3^{x+2}=3^7
    ->x+2=7
    ->x=5
    Vậy x=5
    g,
    9 . 27^x = 3^5
    ->3^2 . 3^{3x} = 3^5
    ->3^{3x+2}=3^5
    ->3x+2=5
    ->3x=3
    ->x=1
    Vậy x=1
    h,
    1/2 . 2^x + 4 . 2^x = 9 . 2^5
    ->2^x . (1/2+4) = 9 . 2^5
    ->2^x . 9/2 = 9 . 2^5
    ->2^x = 2^5 . 9 . 2/9
    ->2^x = 2^5. 2
    ->2^x = 2^6
    ->x=6
    Vậy x=6

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

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