Đáp án:
\(\begin{array}{l}
1)\left\{ \begin{array}{l}
x \ge - 1\\
x \ne 2
\end{array} \right.\\
2)\left\{ \begin{array}{l}
\dfrac{{13}}{7} \le x\\
x \ne \dfrac{5}{2}
\end{array} \right.\\
3)x > 7\\
4)x > \dfrac{5}{2}\\
5)\left\{ \begin{array}{l}
x \ge - \dfrac{1}{2}\\
x < 5
\end{array} \right.\\
6)\dfrac{7}{2} < x\\
7)\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 2 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 1\\
x \ne 2
\end{array} \right.\\
2)DK:\left\{ \begin{array}{l}
13 - 7x \le 0\\
2x - 5 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{13}}{7} \le x\\
x \ne \dfrac{5}{2}
\end{array} \right.\\
3)DK:x - 7 > 0\\
\to x > 7\\
4)DK:2x - 5 > 0\left( {do:{{\left( {x - 2} \right)}^2} \ge 0\forall x} \right)\\
\to x > \dfrac{5}{2}\\
5)DK:\dfrac{{2x + 1}}{{x - 5}} \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 1 \ge 0\\
x - 5 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x + 1 \le 0\\
x - 5 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - \dfrac{1}{2}\\
x < 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - \dfrac{1}{2}\\
x > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
6)\sqrt {\dfrac{{ - \left( {9{x^2} - 6x + 1} \right)}}{{7 - 2x}}} \\
= \sqrt {\dfrac{{ - {{\left( {3x - 1} \right)}^2}}}{{7 - 2x}}} \\
DK:7 - 2x < 0\\
\to \dfrac{7}{2} < x\\
7)DK:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.
\end{array}\)
