75. Let's solve the inequality

$$
\sqrt{2 x+1} \geqslant \frac{x+1}{\sqrt{1-x}}
$$
Solution. a) \( f(x) = \sqrt{2 x + 1} - \frac{x + 1}{\sqrt{1 - x}} \).

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-48.jpg?height=140&width=440&top_left_y=2169&top_left_x=360)

Fig. 40

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-48.jpg?height=157&width=397&top_left_y=2169&top_left_x=838)

Fig. 41

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-48.jpg?height=160&width=460&top_left_y=2184&top_left_x=1255)

Fig. 42

47

Let's find the domain:

\[
\begin{aligned}
& \left(\left\{\begin{array}{l}
2 x + 1 \geqslant 0, \\
1 - x > 0
\end{array}\right) \Leftrightarrow \binom{\left(x \geqslant -\frac{1}{2},\right.}{x < 1} \Leftrightarrow \left(-\frac{1}{2} \leqslant x < 1\right)\right. \text {, i.e. } \\
& D(f) = \left[-\frac{1}{2} ; 1\right) .
\end{aligned}
\]

b) Solve the equation \(\sqrt{2 x + 1} = \frac{x + 1}{\sqrt{1 - x}}\). We have:

\[
(2 x + 1)(1 - x) = (x + 1)^{2} \text {, or } 3 x^{2} + x = 0 \text {, or } x(3 x + 1) = 0 \text { . }
\]

From which \(x_{1} = 0, x_{2} = -\frac{1}{3}\).

By checking, we confirm that both roots satisfy the equation (there are no extraneous roots). Therefore, the function \(f(x)\) has two zeros: \(x_{1} = 0\) and \(x_{2} = -\frac{1}{3}\).

c) Mark \(x_{1} = 0\) and \(x_{2} = -\frac{1}{3}\) on the coordinate line (Fig. 42).

If \(x \in \left[-\frac{1}{2} ; -\frac{1}{3}\right]\), then \(f(x) \leqslant 0\); if \(x \in \left[-\frac{1}{3} ; 0\right)\), then \(f(x) \geqslant 0\); if \(x \in [0 ; 1)\), then \(f(x) \leqslant 0\). Answer: \(\left[-\frac{1}{3} ; 0\right]\).