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:

\(\begin{array}{l}
D = \sqrt {\dfrac{{1 + x}}{{2x – 3}}} \\
DK:\left\{ \begin{array}{l}
\dfrac{{1 + x}}{{2x – 3}} \ge 0\\
2x – 3 \ne 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
2x – 3 < 0
\end{array} \right.
\end{array} \right.\\
x \ne \dfrac{3}{2}
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge  – 1\\
x > \dfrac{3}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le  – 1\\
x < \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x > \dfrac{3}{2}\\
x \le  – 1
\end{array} \right.\\
E = \sqrt {3x – 5}  + \dfrac{2}{{\sqrt {x – 4} }}\\
DK:\left\{ \begin{array}{l}
3x – 5 \ge 0\\
x – 4 > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x \ge \dfrac{5}{3}\\
x > 4
\end{array} \right.\\
 \to x > 4
\end{array}\)