Đáp án:
\(\begin{array}{l}
a)x > 16\\
d)0 < x < 4\\
b)0 \le x < 9\\
e)x \ne 9;x \ge 0\\
c)x \ge 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 16\\
\dfrac{{\sqrt x + 1}}{{\sqrt x - 4}} > 0\\
\to \sqrt x - 4 > 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to x > 16\\
d)DK:x \ge 0;x \ne 4\\
\dfrac{{2\sqrt x + 5}}{{\sqrt x - 2}} \le 1\\
\to \dfrac{{2\sqrt x + 5 - \sqrt x + 2}}{{\sqrt x - 2}} \le 0\\
\to \dfrac{{\sqrt x + 7}}{{\sqrt x - 2}} \le 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x + 7 > 0\forall x \ge 0} \right)\\
\to 0 < x < 4\\
b)DK:x \ge 0;x \ne 9\\
\dfrac{{\sqrt x }}{{\sqrt x - 3}} \le 0\\
\to \sqrt x - 3 < 0\left( {do:\sqrt x \ge 0\forall x \ge 0} \right)\\
\to 0 \le x < 9\\
e)DK:x \ne 9;x \ge 0\\
\dfrac{{2\sqrt x - 5}}{{\sqrt x - 3}} > \dfrac{1}{{\sqrt x - 3}}\\
\to \dfrac{{2\sqrt x - 6}}{{\sqrt x - 3}} > 0\\
\to \dfrac{{2\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}} > 0\\
\to 2 > 0\left( {ld} \right)\\
\to x \ne 9;x \ge 0\\
c)DK:x > 0\\
\dfrac{{\sqrt x - 1}}{{\sqrt x }} \ge 0\\
\to \sqrt x - 1 \ge 0\\
\to x \ge 1
\end{array}\)
