giải bất phương trình 1/(x-2) -1/x-2/(x+2)<=0 23/07/2021 Bởi Elliana giải bất phương trình 1/(x-2) -1/x-2/(x+2)<=0
Đáp án:$\left[ \begin{array}{l} – 1 \le x < 2\\x < – 2\end{array} \right.$ Giải thích các bước giải: $\begin{array}{l}\frac{1}{{x – 2}} – \frac{1}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\ \Rightarrow \frac{{x + 2 – 1}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\ \Rightarrow \frac{{x + 1}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 1 \ge 0\\\left( {x – 2} \right)\left( {x + 2} \right) < 0\end{array} \right.\\\left\{ \begin{array}{l}x + 1 \le 0\\\left( {x – 2} \right)\left( {x + 2} \right) > 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge – 1\\ – 2 < x < 2\end{array} \right.\\\left\{ \begin{array}{l}x \le – 1\\\left[ \begin{array}{l}x > 2\\x < – 2\end{array} \right.\end{array} \right.\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} – 1 \le x < 2\\x < – 2\end{array} \right.\end{array}$ Bình luận
Đáp án:$\left[ \begin{array}{l}
– 1 \le x < 2\\
x < – 2
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\frac{1}{{x – 2}} – \frac{1}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \frac{{x + 2 – 1}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \frac{{x + 1}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
\left( {x – 2} \right)\left( {x + 2} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
\left( {x – 2} \right)\left( {x + 2} \right) > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge – 1\\
– 2 < x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le – 1\\
\left[ \begin{array}{l}
x > 2\\
x < – 2
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
– 1 \le x < 2\\
x < – 2
\end{array} \right.
\end{array}$