Đáp án:
g) \(TXD:D = \left[ { - 3;\dfrac{5}{2}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:{x^2} + 2x - 3 \ne 0\\
\to x \ne \left\{ { - 3;1} \right\}\\
\to TXD:D = R\backslash \left\{ { - 3;1} \right\}\\
b)DK:x - 4 > 0\\
\to x > 4\\
\to TXD:D = \left( {4; + \infty } \right)\\
c)DK:\left\{ \begin{array}{l}
2x + 1 \ge 0\\
3 - x \ge 0
\end{array} \right.\\
\to 3 \ge x \ge - \dfrac{1}{2}\\
\to TXD:D = \left[ { - 3; - \dfrac{1}{2}} \right]\\
d)DK:\left\{ \begin{array}{l}
x - 1 \ne 0\\
3 - x > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 1\\
3 > x
\end{array} \right.\\
\to TXD:D = \left( { - \infty ;3} \right)\backslash \left\{ 1 \right\}\\
e)DK:\left\{ \begin{array}{l}
x - 2 \ne 0\\
7 - 2x \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
\dfrac{7}{2} \ge x
\end{array} \right.\\
\to TXD:D = \left( { - \infty ;\dfrac{7}{2}} \right]\backslash \left\{ 2 \right\}\\
f)DK:\left\{ \begin{array}{l}
2 - 3x \ge 0\\
1 - 2x > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{2}{3} \ge x\\
\dfrac{1}{2} > x
\end{array} \right.\\
\to \dfrac{1}{2} > x\\
TXD:D = \left( { - \infty ;\dfrac{1}{2}} \right)\\
g)DK:\left\{ \begin{array}{l}
x + 3 \ge 0\\
{x^2} - 16 \ne 0\\
5 - 2x > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{2} > x \ge - 3\\
x \ne \pm 4
\end{array} \right.\\
\to TXD:D = \left[ { - 3;\dfrac{5}{2}} \right)
\end{array}\)
