Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.\\
b)x \ge 3\\
c)x \ge 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left( {x + 1} \right)\left( {x - 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
x - 3 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.\\
b)DK:\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 3 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 1\\
x \ge 3
\end{array} \right.\\
\to x \ge 3\\
c)A = B\\
\to \sqrt {\left( {x + 1} \right)\left( {x - 3} \right)} = \sqrt {x + 1} .\sqrt {x - 3} \\
\to \left( {x + 1} \right)\left( {x - 3} \right) = \left( {x + 1} \right).\left( {x - 3} \right)\left( {ld} \right)\forall x \ge 3\\
KL:x \ge 3
\end{array}\)
