Tìm các nghiệm của phương trình: \(x^{3}-15x^{2}+71x-105=0\)  (1)
Biết rằng các nghiệm này lập thành một cấp số cộng \(a-d;a;a+d\) với \(d\neq 0\).
Ta có: \((a-d)^{3}-15(a-d)^{2}+71(a-d)-105=0\)
\(a^{3}-15a^{2}+71a-105=0\)
\((a+d)^{3}-15(a+d)^{2}+71(a+d)-105=0\).
Suy ra: \(6ad^{2}-30d^{2}=d^{2}(6a-30)=0\).
Vì \(d\neq 0\) ta có: \(6a-30=0\Rightarrow a=5\).
Vì \(a=5\) là một nghiệm nên vế trái của (1) chia hết cho \((x-a)\). Do đó:
(1) \(\Leftrightarrow (x-5)(x^{2}-10x+21)=0\Leftrightarrow x=3;x=5;x=7\).
cho mình hỏi làm thế nào suy ra 6ad2−30d2=d2(6a−30)=0 –  totrongha01 29-01-18 09:13 PM

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