Tìm x biết: a) x^5 + x^4 + x + 1 = 0; b) x^4 + 3x^3 – x – 3 = 0; c) x^3 – 5x^2 – x + 5 = 0; d) x(x – 5) – 4x + 20 = 0.
Tìm x biết:
a) x5 + x4 + x + 1 = 0;
b) x4 + 3x3 – x – 3 = 0;
c) x3 – 5x2 – x + 5 = 0;
d) x(x – 5) – 4x + 20 = 0.
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}.