Đáp án:
\(x \in \left( {1;2} \right] \cup \left[ {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge \dfrac{1}{2};x \ne 1\\
Xét:\dfrac{{\left( {\sqrt {x + 1} - \sqrt {2x - 1} } \right)\left( {\sqrt {x + 1} - 2} \right)}}{{x - 1}} = 0\\
\to \left( {\sqrt {x + 1} - \sqrt {2x - 1} } \right)\left( {\sqrt {x + 1} - 2} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt {x + 1} - \sqrt {2x - 1} = 0\\
\sqrt {x + 1} - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + 1 = 2x - 1\\
x + 1 = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.
\end{array}\)
BXD:
x 1/2 1 2 3 +∞
f(x) / + // - 0 + 0 -
\(KL:x \in \left( {1;2} \right] \cup \left[ {3; + \infty } \right)\)
