Toán 7(x) =√4x+1-√-2+1 7(x) 2x+1phan (2x+1)(x-3) đề bài tìm tập xác định 14/09/2021 By Gabriella 7(x) =√4x+1-√-2+1 7(x) 2x+1phan (2x+1)(x-3) đề bài tìm tập xác định
$\begin{array}{l} f\left( x \right) = \sqrt {4x + 1} – \sqrt { – 2x + 1} \\ DK:\left\{ \begin{array}{l} 4x + 1 \ge 0\\ – 2x + 1 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge – \frac{1}{4}\\ x \le \frac{1}{2} \end{array} \right. \Leftrightarrow – \frac{1}{4} \le x \le \frac{1}{2}\\ \Rightarrow TXD:D = \left[ { – \frac{1}{4};\frac{1}{2}} \right]\\ f\left( x \right) = \frac{{2x + 1}}{{\left( {2x + 1} \right)\left( {x – 3} \right)}}\\ DK:\left\{ \begin{array}{l} 2x + 1 \ne 0\\ x – 3 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne – \frac{1}{2}\\ x \ne 3 \end{array} \right.\\ \Rightarrow TXD:D = R\backslash \left\{ { – \frac{1}{2};3} \right\} \end{array}$ Trả lời
$\begin{array}{l}
f\left( x \right) = \sqrt {4x + 1} – \sqrt { – 2x + 1} \\
DK:\left\{ \begin{array}{l}
4x + 1 \ge 0\\
– 2x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge – \frac{1}{4}\\
x \le \frac{1}{2}
\end{array} \right. \Leftrightarrow – \frac{1}{4} \le x \le \frac{1}{2}\\
\Rightarrow TXD:D = \left[ { – \frac{1}{4};\frac{1}{2}} \right]\\
f\left( x \right) = \frac{{2x + 1}}{{\left( {2x + 1} \right)\left( {x – 3} \right)}}\\
DK:\left\{ \begin{array}{l}
2x + 1 \ne 0\\
x – 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne – \frac{1}{2}\\
x \ne 3
\end{array} \right.\\
\Rightarrow TXD:D = R\backslash \left\{ { – \frac{1}{2};3} \right\}
\end{array}$