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

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1. $\\$

   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

   Trả lời
2. $\quad b^2 = ac\qquad (b+c\ne 0)$

   $\Leftrightarrow \dfrac{a}{b} = \dfrac{b}{c}$

   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 $\dfrac{(a+b)^{2021}}{(b+c)^{2021}} = \dfrac{a^{2021} + b^{2021}}{b^{2021} + c^{2021}}$

   Trả lời