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

Toán Lớp 7: Chứng minh rằng nếu `a/b=c/d` thì: `{5a+3b}/{5a-3b}={5c+3d}/{5c-3d}` `{7a^2+3ab}/{11a^2-8b^2}={7c^2+3cd}/{11c^2-8d^2}`

Tháng Tám 4, 2022 Toán học 2 Comments 0 views

Toán Lớp 7: Chứng minh rằng nếu a/b=c/d thì:
{5a+3b}/{5a-3b}={5c+3d}/{5c-3d}
{7a^2+3ab}/{11a^2-8b^2}={7c^2+3cd}/{11c^2-8d^2}

Comments ( 2 )

  1. cobexinhdep
    cobexinhdep
    4 Tháng Tám, 2022 at 5:13 chiều
    Reply
    Gửi bạn
     

    toan-lop-7-chung-minh-rang-neu-a-b-c-d-thi-5a-3b-5a-3b-5c-3d-5c-3d-7a-2-3ab-11a-2-8b-2-7c-2-3cd

  2. cattien
    cattien
    4 Tháng Tám, 2022 at 5:12 chiều
    Reply
    $\\$
    Đặt a/b=c/d=k(k\ne 0)
    ->a=bk,c=dk
    a,
    (5a+3b)/(5a-3b)=(5bk+3b)/(5bk-3b)=(b(5k+3))/(b(5k-3))=(5k+3)/(5k-3)
    (5c+3d)/(5c-3d)=(5dk+3d)/(5dk-3d)=(d(5k+3))/(d(5k-3))=(5k+3)/(5k-3)
    -> (5a+3b)/(5a-3b)=(5c+3d)/(5c-3d)(=(5k+3)/(5k-3))
    b,
    (7a^2+3ab)/(11a^2 – 8b^2)
    = (7b^2k^2 + 3b^2k)/(11b^2k^2 – 8b^2)
    = (b^2 (7k^2 + 3k))/(b^2 (11k^2 – 8))
    = (7k^2 +3k)/(11k^2-8) (1)
    (7c^2+3cd)/(11c^2 – 8d^2)
    =(7d^2k^2 +3d^2k)/(11d^2k^2 – 8d^2)
    = (d^2 (7k^2 +3k))/(d^2(11k^2 – 8))
    = (7k^2+3k)/(11k^2 – 8) (2)
    Từ (1)(2)
    -> (7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)(=(7k^2+3k)/(11k^2-8))

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

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