Toán Lớp 9: Tìm điều kiện làm từng bước giúp mình √-3x/x ²-1
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Toán Lớp 9: Tìm điều kiện làm từng bước giúp mình √-3x/x ²-1
Comments ( 1 )
\sqrt {\dfrac{{ – 3x}}{{{x^2} – 1}}} \\
Dkxd:\dfrac{{ – 3x}}{{{x^2} – 1}} \ge 0\\
\Leftrightarrow \dfrac{{3x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 0\\
\left( {x – 1} \right)\left( {x + 1} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 0\\
\left( {x – 1} \right)\left( {x + 1} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 0\\
\left[ \begin{array}{l}
x > 1\\
x < – 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 0\\
– 1 < x < 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < – 1\\
0 \le x < 1
\end{array} \right.\\
Vay\,x < – 1\,hoặc\,0 \le x < 1
\end{array}$