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

Toán Lớp 7: Cho b^2=ac (b+c khác 0) Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$

Home/ Toán Lớp 7: Cho b^2=ac (b+c khác 0) Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$

Toán Lớp 7: Cho b^2=ac (b+c khác 0) Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$

Tháng Tư 26, 2023 Toán học 2 Comments 0 views

Toán Lớp 7: Cho b^2=ac (b+c khác 0)
Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$

Comments ( 2 )

  1. landinhngoc97
    landinhngoc97
    26 Tháng Tư, 2023 at 6:20 chiều
    Reply
    $\quad b^2 = ac\qquad (b+c\ne 0)$
    $\Leftrightarrow \dfrac{a}{b} = \dfrac{b}{c}$
    Đặt $\dfrac{a}{b} = \dfrac{b}{c} = k$
    $\Rightarrow \begin{cases}a = kb\\b = kc\end{cases}$
    Ta được:
    $\bullet\quad \dfrac{(a+b)^{2021}}{(b+c)^{2021}}$
    $= \dfrac{(kb + b)^{2021}}{(kc + c)^{2021}}$
    $= \dfrac{b^{2021}(k+1)^{2021}}{c^{2021}(k+1)^{2021}}$
    $= \dfrac{b^{2021}}{c^{2021}}$
    $\bullet\quad \dfrac{a^{2021} + b^{2021}}{b^{2021} + c^{2021}}$
    $= \dfrac{(kb)^{2021} + b^{2021}}{(kc)^{2021} + c^{2021}}$
    $= \dfrac{b^{2021}(k^{2021} + 1)}{c^{2021}(k^{2021} + 1)}$
    $= \dfrac{b^{2021}}{c^{2021}}$
    Vậy $\dfrac{(a+b)^{2021}}{(b+c)^{2021}} = \dfrac{a^{2021} + b^{2021}}{b^{2021} + c^{2021}}$

  2. dinhcongson
    dinhcongson
    26 Tháng Tư, 2023 at 6:20 chiều
    Reply
    $\\$
    b^2=ac(b+c\ne 0)
    <=>b.b=a.c
    <=> b/a=c/b (*)
    Đặt (*) =k(k\ne 0)
    <=>b=ak, c=bk
    (a+b)^{2021}/(b+c)^{2021}
    =(a+ak)^{2021}/(b+bk)^{2021}
    =(a(1+k))^{2021}/(b(1+k))^{2021}
    =(a^{2021}(1+k)^{2021})/(b^{2021}(1+k)^{2021})
    =a^{2021}/b^{2021}(1)
    (a^{2021}+b^{2021})/(b^{2021}+c^{2021})
    =(a^{2021}+a^{2021}k^{2021})/(b^{2021}+b^{2021}k^{2021})
    =(a^{2021}(1+k^{2021}))/(b^{2021}(1+k^{2021}))
    =a^{2021}/b^{2021}(2)
    (1)(2)=>đpcm

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

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