Đáp án:
\(\begin{array}{l}
6)TXD:D = \left[ { - 2; + \infty } \right)\backslash \left\{ { - 1} \right\}\\
7)TXD:D = \left( { - \infty ;\dfrac{7}{3}} \right)\\
8)TXD:D = R\backslash \left\{ {1;2} \right\}\\
9)TXD:D = \left[ {1; + \infty } \right)\backslash \left\{ 5 \right\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)DK:\left\{ \begin{array}{l}
x + 2 \ge 0\\
\sqrt {x + 2} - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 2\\
x + 2 \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 2\\
x \ne - 1
\end{array} \right.\\
\to TXD:D = \left[ { - 2; + \infty } \right)\backslash \left\{ { - 1} \right\}\\
7)DK:7 - 3x > 0\\
\to \dfrac{7}{3} > x\\
\to TXD:D = \left( { - \infty ;\dfrac{7}{3}} \right)\\
8)DK:{x^2} - 3x + 2 \ne 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) \ne 0\\
\to x \ne \left\{ {1;2} \right\}\\
\to TXD:D = R\backslash \left\{ {1;2} \right\}\\
9)DK:\left\{ \begin{array}{l}
x - 1 \ge 0\\
2 - \sqrt {x - 1} \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 1\\
4 \ne x - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 1\\
x \ne 5
\end{array} \right.\\
\to TXD:D = \left[ {1; + \infty } \right)\backslash \left\{ 5 \right\}
\end{array}\)
