Question

x^5=4x^4+4x^3+4x^2+4x+5

Answer 1

\(PT\Leftrightarrow x^5-1=4\left(x^4+x^3+x^2+x+1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x^4+x^3+x^2+x+1=0\end{matrix}\right.\).

Nếu \(x^4+x^3+x^2+x+1=0\Rightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=0\Leftrightarrow x^5-1=0\Leftrightarrow x^5=1\Leftrightarrow x=1\). Thử lại ta thấy không thoả mãn.

Do đó ta có \(x-1=4\Leftrightarrow x=5\).

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