e 7 còn c?

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a;(- \(x + 5\)).(3 - \(x\)) = 0

\(\left[\right. - x + 5 = 0 \\ 3 - x = 0\)

\(\left[\right. x = 5 \\ x = 3\)

Vậy \(x \in\) {3; 5}

b; (\(x - 1\)) x (\(x + 2\)) x (- \(x - 3\)) = 0

\(\left[\right. x - 1 = 0 \\ x + 2 = 0 \\ - x - 3 = 0\)

\(\left[\right. x = 1 \\ x = - 2 \\ x = - 3\)

Vậy \(x\) ∈ {-2; -3; 1}

T_T

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