Lời giải

a) x⁵ + x⁴ + x + 1 = 0

⇔ x⁴(x + 1) + (x + 1) = 0

⇔ (x + 1)(x⁴ + 1) = 0

⇔ x + 1 = 0 (vì x⁴ + 1 > 0)

⇔ x = – 1

Vậy x = – 1.

b) x⁴ + 3x³ – x – 3 = 0

⇔ x³(x + 3) – (x + 1) = 0

⇔ (x + 3)(x³ – 1) = 0

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

\( \Leftrightarrow \left[ \begin{array}{l}x = - 3\\x = 1\end{array} \right.\)

Vậy x ∈ {– 3; 1}.

c) x³ – 5x² – x + 5 = 0

⇔ x(x² – 1) – 5(x² – 1) = 0

⇔ (x² – 1)(x – 5) = 0

\( \Leftrightarrow \left[ \begin{array}{l}{x^2} - 1 = 0\\x - 5 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = 1\\x = 5\end{array} \right.\)

Vậy x ∈ {1; 5; – 1}.

d) x(x – 5) – 4x + 20 = 0

⇔ x(x – 5) – 4(x – 5)= 0

⇔ (x – 4)(x – 5) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x - 4 = 0\\x - 5 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 4\\x = 5\end{array} \right.\)

Vậy x ∈ {4; 5}.