Question

Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)

CMR :(\(\dfrac{a+b+c+d}{b+c+d+e}\))⁴=\(\dfrac{a}{e}\)

giúp mình giải bài này nha

Answer 1

Đặt \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)

Ta có: \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{a+b+c+d}{b+c+d+e}\) ( t/c dãy tỉ số bằng nhau )

\(\Rightarrow k^4=\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4\) (1)

\(k^4=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}=\dfrac{a}{e}\) (2)

Từ (1), (2) \(\Rightarrow\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4=\dfrac{a}{e}\left(đpcm\right)\)

Vậy...

Answer 2

áp dụng tc dãy tỉ ra luôn

Answer 3

Ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)

=> \(\left(\dfrac{a}{b}\right)^4=\dfrac{a.b.c.d}{b.c.d.e}=\dfrac{a}{e}\) (1)

Mà :

\(\dfrac{a}{b}=\dfrac{a+b+c+d}{b+c+d+e}\left(TTCCDTSBN\right)\)

=> \(\left(\dfrac{a}{b}\right)^4=\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4\)(2)

=> từ (1) và (2)

=> \(\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4=\dfrac{a}{e}\left(đp.....\right)\)