\(\left\{{}\begin{matrix}a^2_2=a_1a_3\\a^2_3=a_2a_4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}\\\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}\end{matrix}\right.\Rightarrow\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}\)
Đặt: \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}=t\)
\(\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}.\dfrac{a_3}{a_4}=t.t.t=\dfrac{a_1}{a_4}=t^3\left(1\right)\)
Ta có:\(\left\{{}\begin{matrix}\dfrac{a^3_1}{a^3_2}=t^3\\\dfrac{8a^3_2}{8a^3_3}=t^3\\\dfrac{125a^3_3}{125a^3_4}=t^3\end{matrix}\right.\) \(\Rightarrow\dfrac{a^3_1}{a^3_2}=\dfrac{8a^3_2}{8a^3_3}=\dfrac{125a^3_3}{125a^3_4}=\dfrac{a^3_1+8a^3_2+125a^3_3}{a^3_2+8a^3_3+125a^3_4}=t^3\)
Ta có đpcm
Ta có: \(a_2^2=a_1.a_3\Leftrightarrow a_2.a_2=a_1.a_3\Leftrightarrow\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}\left(1\right)\)
\(a_3^2=a_2.a_4\Leftrightarrow a_3.a_3=a_2.a_4\Leftrightarrow\dfrac{a_2}{a_3}=\dfrac{a_3}{a_4}\left(2\right)\)
Từ
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{a_1^3}{a^3_2}=\dfrac{8a_2^3}{8a^3_3}=\dfrac{125a_3^3}{125a^3_4}=\dfrac{a_1^3+8a_2^3+125a^3_3}{a^3_2+8a^3_3+125a^3_4}\left(3\right)\)
Ta lại có: \(\dfrac{a_1^3}{a^3_2}=\left(\dfrac{a_1}{a_2}\right)^3=\dfrac{a_1}{a_2}\cdot\dfrac{a_1}{a_2}.\dfrac{a_1}{a_2}=\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}.\dfrac{a_3}{a_4}=\dfrac{a_1}{a_4}\left(4\right)\)
Từ \(\left(3\right);\left(4\right)\Rightarrow\dfrac{a_1}{a_4}=\dfrac{a_1^3+8a_2^3+125a_3^3}{a^3_2+8a_3^3+125a^3_4}\left(dpcm\right)\)
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