đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\left(k\ne0\right)\)
vì\(\dfrac{a}{b}=k\Rightarrow a=bk;\dfrac{b}{c}=k\Rightarrow b=ck;\dfrac{c}{d}=k\Rightarrow c=dk;\dfrac{d}{e}=k\Rightarrow d=ek\)
Ta có:
\(\dfrac{a}{e}=\dfrac{kb}{e}=\dfrac{kkc}{e}=\dfrac{kkkd}{e}=\dfrac{kkkke}{e}=k^4\)
\(\left(\dfrac{a-b+c-d}{b-c+d-e}\right)^4=\left(\dfrac{kb-kc+kd-ke}{b-c+d-e}\right)^4=\left(\dfrac{k\left(b-c+d-e\right)}{d-c+d-e}\right)^4=k^4\)vì \(k^4=k^4\Rightarrow\dfrac{a}{e}=\left(\dfrac{a-b+c-d}{b-c+d-e}\right)^4\)
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