Question

Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)và \(a+b+c+d\ne0\)

CMR:\(a^{20}.b^{17}.c^{2017}=d^{2054}\)

Answer 1

Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\)

\(\Rightarrow\dfrac{a}{b}=1\Rightarrow a=b\)

\(\dfrac{b}{c}=1\Rightarrow b=c\)

\(\dfrac{c}{d}=1\Rightarrow c=d\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow a^{20}.b^{17}.c^{2017}=d^{20}.d^{17}.d^{2017}=d^{2054}\)

                                                      đpcm

Tham khảo nhé~

Answer 2

Theo đề bài, ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)

Áp dụng dãy tỉ số bằng nhau, ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=1\Rightarrow a=b\\\frac{b}{c}=1\Rightarrow b=c\\\frac{c}{d}=1\Rightarrow c=d\end{cases}}\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow a^{20}.b^{17}.c^{2017}=d^{20}.d^{17}.d^{2017}=d^{2054}\)

\(\Rightarrowđpcm\)