Đáp án:
B10:
g. \(\left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B9:\\
e.DK:\left\{ \begin{array}{l}
\dfrac{{x - 2}}{{x + 3}} \ge 0\\
x + 3 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne - 3\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 \ge 0\\
x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 \le 0\\
x + 3 < 0
\end{array} \right.
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 2\\
x < - 3
\end{array} \right.\\
g.DK:1 - {x^2} \ge 0\\
\to \left( {1 - x} \right)\left( {1 + x} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - x \ge 0\\
1 + x \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - x \le 0\\
1 + x \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
1 \ge x \ge - 1\\
\left\{ \begin{array}{l}
x \ge 1\\
x \le - 1
\end{array} \right.\left( l \right)
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
B10:\\
a.Do:{x^2} + 1 > 0\forall x \in R\\
\to DK:\forall x\\
b.DK:2x - 1 > 0\\
\to x > \dfrac{1}{2}\\
d.DK:\left\{ \begin{array}{l}
x - 2 \ge 0\\
x - 3 \ge 0
\end{array} \right.\\
\to x \ge 3\\
e.DK:\left( {x - 2} \right)\left( {x + 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 2\\
x \le - 3
\end{array} \right.\\
c.DK:\left\{ \begin{array}{l}
x \ge 0\\
x \ge 1
\end{array} \right.\\
g.DK:\left\{ \begin{array}{l}
\dfrac{{2x - 3}}{{x - 1}} \ge 0\\
x - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.\\
x \ne 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.
\end{array}\)
