1. ( 2 căn x – 1 ) ( 3 căn x -6 ) = 0 2.căn x -2 / căn x – 3 > 0 3.căn x +1 / căn x-2 <0 15/07/2021 Bởi Kaylee 1. ( 2 căn x – 1 ) ( 3 căn x -6 ) = 0 2.căn x -2 / căn x – 3 > 0 3.căn x +1 / căn x-2 <0
Đáp án: 3) \(0 \le x < 4\) Giải thích các bước giải: \(\begin{array}{l}1)DK:x \ge 0\\\left( {2\sqrt x – 1} \right)\left( {3\sqrt x – 6} \right) = 0\\ \to \left[ \begin{array}{l}2\sqrt x – 1 = 0\\3\sqrt x – 6 = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{1}{4}\\\sqrt x = 2\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{1}{4}\\x = 4\end{array} \right.\\2)DK:x \ge 0;x \ne 9\\\dfrac{{\sqrt x – 2}}{{\sqrt x – 3}} > 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}\sqrt x – 2 > 0\\\sqrt x – 3 > 0\end{array} \right.\\\left\{ \begin{array}{l}\sqrt x – 2 < 0\\\sqrt x – 3 < 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 4\\x > 9\end{array} \right.\\\left\{ \begin{array}{l}x < 4\\x < 9\end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}x > 9\\0 \le x < 4\end{array} \right.\\3)DK:x \ge 0;x \ne 4\\\dfrac{{\sqrt x + 1}}{{\sqrt x – 2}} < 0\\ \to \sqrt x – 2 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\ \to 0 \le x < 4\end{array}\) Bình luận
Đáp án:
3) \(0 \le x < 4\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0\\
\left( {2\sqrt x – 1} \right)\left( {3\sqrt x – 6} \right) = 0\\
\to \left[ \begin{array}{l}
2\sqrt x – 1 = 0\\
3\sqrt x – 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 4
\end{array} \right.\\
2)DK:x \ge 0;x \ne 9\\
\dfrac{{\sqrt x – 2}}{{\sqrt x – 3}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x – 2 > 0\\
\sqrt x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt x – 2 < 0\\
\sqrt x – 3 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 4\\
x > 9
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 4\\
x < 9
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 9\\
0 \le x < 4
\end{array} \right.\\
3)DK:x \ge 0;x \ne 4\\
\dfrac{{\sqrt x + 1}}{{\sqrt x – 2}} < 0\\
\to \sqrt x – 2 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 4
\end{array}\)