Đáp án:
$\begin{array}{l}
b4)\\
a)\dfrac{{\sqrt {2x - 7} }}{{\sqrt {x + 4} }}\\
Dkxd:\left\{ \begin{array}{l}
2x - 7 \ge 0\\
x + 4 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{7}{2}\\
x > - 4
\end{array} \right. \Leftrightarrow x \ge \dfrac{7}{2}\\
b)\sqrt {\dfrac{{2x - 7}}{{x + 4}}} \\
Dkxd:\dfrac{{2x - 7}}{{x + 4}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{7}{2}\\
x < - 4
\end{array} \right.\\
Vậy\,x \ge \dfrac{7}{2}\,hoặc\,x < - 4\\
c)\dfrac{{\sqrt {x - 4} }}{{\sqrt { - 2x + 3} }}\\
Dkxd:\left\{ \begin{array}{l}
x - 4 \ge 0\\
- 2x + 3 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
x < \dfrac{3}{2}
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
d)\sqrt {\dfrac{{x - 4}}{{ - 2x + 3}}} \\
Dkxd:\dfrac{{x - 4}}{{ - 2x + 3}} \ge 0 \Leftrightarrow \dfrac{{x - 4}}{{2x - 3}} \le 0\\
\Leftrightarrow \dfrac{3}{2} < x \le 4\\
Vậy\,\dfrac{3}{2} < x \le 4\\
B5)a)\sqrt {{x^2} - 4} \\
Dkxd:{x^2} - 4 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.\\
Vậy\,x \ge 2\,hoặc\,x \le - 2\\
b)\sqrt {9 - 4{x^2}} \\
Dkxd:9 - 4{x^2} \ge 0\\
\Leftrightarrow {x^2} \le \dfrac{9}{4}\\
\Leftrightarrow \dfrac{{ - 3}}{2} \le x \le \dfrac{3}{2}\\
Vậy\, - \dfrac{3}{2} \le x \le \dfrac{3}{2}\\
c)\sqrt {{x^2} - 4x + 3} \\
Dkxd:{x^2} - 4x + 3 \ge 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le 1
\end{array} \right.\\
Vậy\,x \le 1\,hoặc\,x \ge 3\\
d)\sqrt {{x^2} - 4x - 5} \\
Dkxd:{x^2} - 4x - 5 \ge 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x + 1} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 5\\
x \le - 1
\end{array} \right.\\
Vậy\,x \le - 1\,hoặc\,x \ge 5\\
f)\sqrt {{x^2} - 4x + 4} \\
Dkxd:{x^2} - 4x + 4 \ge 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} \ge 0\left( {tm} \right)\\
Vậy\,x \in R
\end{array}$
