Đáp án:
f. \(x \ge \dfrac{7}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\sqrt {2x - 5} \\
DK:2x - 5 \ge 0 \to x \ge \dfrac{5}{2}\\
c.\sqrt {\dfrac{{{x^2} + 1}}{{2x - 3}}} \\
DK:2x - 3 > 0 \to x > \dfrac{3}{2}\\
g.\sqrt { - x + 5} \\
DK:5 - x \ge 0 \to 5 \ge x\\
k.\sqrt {\dfrac{5}{{2x + 4}}} \\
DK:2x + 4 > 0 \to x > - 2\\
c.\sqrt {\dfrac{1}{{x - 2}}} \\
DK:x - 2 > 0 \to x > 2\\
e.\sqrt {\dfrac{{x + 1}}{{{x^2} + 1}}} \\
DK:x + 1 \ge 0\\
\to x \ge - 1\\
h.\sqrt {3 - 4x} \\
DK:3 - 4x \ge 0\\
\to x \le \dfrac{3}{4}\\
l.\sqrt {3x + 11} \\
DK:3x + 11 \ge 0\\
\to x \ge - \dfrac{{11}}{3}\\
l.\sqrt {\dfrac{{3x + 7}}{6}} \\
DK:3x + 7 \ge 0\\
\to x \ge - \dfrac{7}{3}\\
f.\sqrt {3x - 7} \\
DK:3x - 7 \ge 0 \to x \ge \dfrac{7}{3}
\end{array}\)
