Tìm x biết:

a) (x – 5).(x + 4) = 0;

b) x² – 7x = 0;

c) x² = –5x;

d) x³ = x;

e) (x – 5).(x – 4) = 0.

Lời giải

a) (x – 5).(x + 4) = 0.

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

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

Vậy \(x \in \left\{ {5; - 4} \right\}\).

b) x² – 7x = 0.

⇔ x.(x – 7) = 0.

\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x - 7 = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 7\end{array} \right.\)

Vậy \(x \in \left\{ {0;7} \right\}\).

c) x² = –5x.

⇔ x² + 5x = 0.

⇔ x.(x + 5) = 0.

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

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

Vậy \(x \in \left\{ {0; - 5} \right\}\).

d) x³ = x.

⇔ x³ – x = 0.

⇔ x.(x² – 1) = 0.

⇔ x.(x – 1)(x + 1) = 0.

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

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

Vậy \(x \in \left\{ {0;1; - 1} \right\}\).

e) (x – 5).(x – 4) = 0.

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

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

Vậy \(x \in \left\{ {5;4} \right\}\).