Tìm đkxđ D = Căn tất cả 1+×/2x-3 E= Căn 3x-5 + 2/ căn x-4 19/07/2021 Bởi Adeline Tìm đkxđ D = Căn tất cả 1+×/2x-3 E= Căn 3x-5 + 2/ căn x-4
Đáp án: e. x>4 Giải thích các bước giải: \(\begin{array}{l}D = \sqrt {\dfrac{{1 + x}}{{2x – 3}}} \\DK:\left\{ \begin{array}{l}\dfrac{{1 + x}}{{2x – 3}} \ge 0\\2x – 3 \ne 0\end{array} \right.\\ \to \left\{ \begin{array}{l}\left[ \begin{array}{l}\left\{ \begin{array}{l}x + 1 \ge 0\\2x – 3 > 0\end{array} \right.\\\left\{ \begin{array}{l}x + 1 \le 0\\2x – 3 < 0\end{array} \right.\end{array} \right.\\x \ne \dfrac{3}{2}\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge – 1\\x > \dfrac{3}{2}\end{array} \right.\\\left\{ \begin{array}{l}x \le – 1\\x < \dfrac{3}{2}\end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}x > \dfrac{3}{2}\\x \le – 1\end{array} \right.\\E = \sqrt {3x – 5} + \dfrac{2}{{\sqrt {x – 4} }}\\DK:\left\{ \begin{array}{l}3x – 5 \ge 0\\x – 4 > 0\end{array} \right.\\ \to \left\{ \begin{array}{l}x \ge \dfrac{5}{3}\\x > 4\end{array} \right.\\ \to x > 4\end{array}\) Bình luận
Đáp án:
e. x>4
Giải thích các bước giải:
\(\begin{array}{l}
D = \sqrt {\dfrac{{1 + x}}{{2x – 3}}} \\
DK:\left\{ \begin{array}{l}
\dfrac{{1 + x}}{{2x – 3}} \ge 0\\
2x – 3 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
2x – 3 < 0
\end{array} \right.
\end{array} \right.\\
x \ne \dfrac{3}{2}
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge – 1\\
x > \dfrac{3}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le – 1\\
x < \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{3}{2}\\
x \le – 1
\end{array} \right.\\
E = \sqrt {3x – 5} + \dfrac{2}{{\sqrt {x – 4} }}\\
DK:\left\{ \begin{array}{l}
3x – 5 \ge 0\\
x – 4 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge \dfrac{5}{3}\\
x > 4
\end{array} \right.\\
\to x > 4
\end{array}\)