The image appears to present several math problems related to limits, derivatives, and inequalities. It defines a function $y=f(x) = -x - \frac{4}{x}$. The problems seem to involve: 1. Calculating the limits of the function $f(x)$ as $x$ approaches $0$ from the right and as $x$ approaches infinity.

I am unable to fully understand and solve the problem due to the image quality and the presence of a language other than English. However, I can provide some initial analysis based on the readable parts.

1. Problem Description

The image appears to present several math problems related to limits, derivatives, and inequalities. It defines a function $y=f(x) = -x - \frac{4}{x}$.

The problems seem to involve:

1. Calculating the limits of the function $f(x)$ as $x$ approaches $0$ from the right and as $x$ approaches infinity.

2. Finding the equation of a line $L: y=x$.

3. Analyzing the inequality $x^2 + 4 - 4\ln(x) > 0$ for $x > 0$ and $f'(x) < 0$.

2. Solution Steps

Because of the unclear text in the image, here are some general approaches to solving these types of problems.

1. Calculating Limits:

- The limit as $x \to 0^+$ of $f(x) = -x - \frac{4}{x}$:

$\lim_{x\to 0^+} (-x - \frac{4}{x}) = -0 - \frac{4}{0^+} = -\infty$.

- The limit as $x \to +\infty$ of $f(x) = -x - \frac{4}{x}$:

$\lim_{x\to +\infty} (-x - \frac{4}{x}) = -\infty - 0 = -\infty$.

2. Equation of Line:

- The equation of the line $L$ is given as $y = x$.

- Need more information to understand what is being asked about this line.

3. Inequality and Derivative Analysis:

- We're given $x^2 + 4 - 4\ln(x) > 0$ for $x>0$. This involves analyzing the behavior of the function $g(x) = x^2 + 4 - 4\ln(x)$.

- To analyze the inequality, we can examine the derivative $g'(x) = 2x - \frac{4}{x} = \frac{2x^2 - 4}{x}$. Setting $g'(x) = 0$, we get $x^2 = 2$, so $x = \sqrt{2}$ (since $x>0$).

- To check if $f'(x) < 0$, we first need to calculate $f'(x)$ for $f(x) = -x - \frac{4}{x}$.

$f'(x) = -1 + \frac{4}{x^2}$.

- Solving the inequality $f'(x) < 0$, we get $-1 + \frac{4}{x^2} < 0$, which implies $\frac{4}{x^2} < 1$, so $x^2 > 4$. Since $x > 0$, we have $x > 2$.

3. Final Answer

Due to the unclear nature of the original problem and image quality, I cannot provide a complete, final answer.

However, here are the calculations based on what I can read:

- $\lim_{x\to 0^+} (-x - \frac{4}{x}) = -\infty$

- $\lim_{x\to +\infty} (-x - \frac{4}{x}) = -\infty$

- $L: y = x$

- $f'(x) < 0$ when $x>2$.