Given the inequality:
$$\sqrt{\frac{x + 1}{2 x - 1}} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{\frac{x + 1}{2 x - 1}} = 1$$
Solve:
$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\sqrt{\frac{x + 1}{2 x - 1}} < 1$$
$$\sqrt{\frac{1 + \frac{19}{10}}{-1 + \frac{2 \cdot 19}{10}}} < 1$$
_____
\/ 203
------- < 1
14
but
_____
\/ 203
------- > 1
14
Then
$$x < 2$$
no execute
the solution of our inequality is:
$$x > 2$$
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/
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x1