Question

cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\) và \(a+b+c\ne0\)

tính \(\dfrac{a^3+b^2+c^{1930}}{a^{1935}}\)

Answer 1

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{c}{a}=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)\(\Rightarrow a=b=c\)

Khi đó \(A=\dfrac{a^3b^2c^{1930}}{a^{1935}}=\dfrac{a^3\cdot a^2\cdot a^{1930}}{a^{1935}}=\dfrac{a^{1935}}{a^{1935}}=1\)