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

Toán Lớp 7: Cho `b^2=ac`;`c^2=bd`. Chứng minh rằng `{a^3+b^3-c^3}/{b^3+c^3-d^3}=({a+b-c}/{b+c-d})^3` `a/b={a^3+8b^3+125c^3}/{b^3+8c^3+125d^3}`

Tháng Tám 4, 2022 Toán học 1 Comment 0 views

Toán Lớp 7: Cho b^2=ac;c^2=bd. Chứng minh rằng
{a^3+b^3-c^3}/{b^3+c^3-d^3}=({a+b-c}/{b+c-d})^3
a/b={a^3+8b^3+125c^3}/{b^3+8c^3+125d^3}

Comments ( 1 )

  1. hongocha12
    hongocha12
    4 Tháng Tám, 2022 at 5:01 chiều
    Reply
    $\\$
    b^2=ac
    ->b.b=a.c
    ->b/a=c/b (1)
    c^2=bd
    ->c.c=b.d
    ->c/b=d/c (2)
    (1)(2)
    ->b/a=c/b=d/c
    Đặt b/a=c/b=d/c=k(k\ne 0)
    ->b=ak, c=bk,d=ck
    $\\$
    a,
    (a^3+b^3-c^3)/(b^3+c^3-d^3)
    = (a^3+b^3-c^3)/(a^3k^3 +b^3k^3 – c^3k^3)
    =(a^3 +b^3-c^3)/(k^3(a^3+b^3-c^3))
    = 1/k^3 
    ((a+b-c)/(b+c-d))^3
    =((a+b-c)/(ak+bk-ck))^3
    =((a+b-c)/(k(a+b-c)))^3
    = 1/k^3
    Do đó : (a^3+b^3-c^3)/(b^3+c^3-d^3) =((a+b-c)/(b+c-d))^3(=1/k^3)
    $\\$
    b,
    a/d
    =(a.b.c)/(b.c.d)
    =(a.b.c)/(ak.bk.ck)
    =(a.b.c)/(a.b.c.k^3)
    =1/k^3
    (a^3 + 8b^3 +125c^3)/(b^3+8c^3+125d^3)
    =(a^3+8b^3+125c^3)/(a^3k^3 + 8b^3k^3 + 125c^3k^3)
    =(a^3+8b^3+125c^3)/(k^3(a^3+8b^3+125c^3))
    =1/k^3
    Do đó : a/d=(a^3 + 8b^3 +125c^3)/(b^3+8c^3+125d^3)(=1/k^3)   

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

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