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