[Toán 7] Tìm n

D

dominhnhut23

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N

nhuquynhdat

Từ $b^2=ac \Longrightarrow \dfrac{a}{b}=\dfrac{b}{c}$

$\Longrightarrow \dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}$

Mặt khác, từ $\dfrac{a}{b}=\dfrac{b}{c} =\dfrac{2014b}{2014c}=\dfrac{a+2014b}{b+2014c}$

$\Longrightarrow \dfrac{a^2}{b^2}=(\dfrac{a+2014b}{b+2014c})^2$

$\Longrightarrow \dfrac{a}{c}=(\dfrac{a+2014b}{b+2014c})^2$

Mà $\dfrac{a}{c}=(\dfrac{a+2014b}{b+2014c})^n \Longrightarrow (\dfrac{a+2014b}{b+2014c})^2=(\dfrac{a+2014b}{b+2014c})^n \Longrightarrow n=2$
 
T

thieukhang61

Từ $b^2=ac \Longrightarrow \dfrac{a}{b}=\dfrac{b}{c}$

$\Longrightarrow \dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}$

Mặt khác, từ $\dfrac{a}{b}=\dfrac{b}{c} =\dfrac{2014b}{2014c}=\dfrac{a+2014b}{b+2014c}$

$\Longrightarrow \dfrac{a^2}{b^2}=(\dfrac{a+2014b}{b+2014c})^2$

$\Longrightarrow \dfrac{a}{c}=(\dfrac{a+2014b}{b+2014c})^2$

Mà $\dfrac{a}{c}=(\dfrac{a+2014b}{b+2014c})^n \Longrightarrow (\dfrac{a+2014b}{b+2014c})^2=(\dfrac{a+2014b}{b+2014c})^n \Longrightarrow n=2$
n=1 hay n=0 cũng thỏa mãn mà bạn. Theo mình đầu tiên nên xét a khác b khác c và a=b=c trước.
 
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