Đáp án:
\(\begin{array}{l}
a,\\
x > 2\\
b,\\
x \ge 2\\
c,\\
x > 2\\
d,\\
x < \dfrac{3}{2}\\
e,\\
x > - \dfrac{3}{2}\\
f,\\
x < - 1
\end{array}\)
Giải thích các bước giải:
Mỗi biểu thức đã cho có nghĩa khi:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
x - 2 \ne 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ge 2
\end{array} \right. \Leftrightarrow x > 2\\
b,\\
\left\{ \begin{array}{l}
x + 2 \ne 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ge 2
\end{array} \right. \Leftrightarrow x \ge 2\\
c,\\
\left\{ \begin{array}{l}
{x^2} - 4 \ne 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ne 4\\
x \ge 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 2\\
x \ge 2
\end{array} \right. \Leftrightarrow x > 2\\
d,\\
\left\{ \begin{array}{l}
\dfrac{1}{{3 - 2x}} \ge 0\\
3 - 2x \ne 0
\end{array} \right. \Leftrightarrow 3 - 2x > 0 \Leftrightarrow 2x < 3 \Leftrightarrow x < \dfrac{3}{2}\\
e,\\
\left\{ \begin{array}{l}
\dfrac{4}{{2x + 3}} \ge 0\\
2x + 3 \ne 0
\end{array} \right. \Leftrightarrow 2x + 3 > 0 \Leftrightarrow 2x > - 3 \Leftrightarrow x > - \dfrac{3}{2}\\
f,\\
\left\{ \begin{array}{l}
\dfrac{{ - 2}}{{x + 1}} \ge 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow x + 1 < 0 \Leftrightarrow x < - 1
\end{array}\)
