Question

(1/(x-1))+(3/(|x|+1))>=(1/(|x|-1))

Answer 1

(1/(x-1))+(3/(|x|+1))>=(1/(|x|-1))

ОДЗ x≠ 1 x≠-1

1. x>=0

(1/(x-1))+(3/(x+1))>=(1/(x-1))

(3/(x+1))>= 0

для всех x>0 кроме 1

2. x<0

1/(x-1) + 3/(-x+1) >= 1/(-x-1)

2/(1 - x) + 1/(x+1) >= 0

(2x+ 2 + 1-x)/(1-x)(1+x) >=0

(x + 3)/(1-x)(1+x) >=0

(x+3)(x-1)(x+1) <=0

-----------[-3] ++++++++ (-1) ------------ (1) ++++++++++

x ∈ (-∞  -3] U (-1  1)     x<0

x∈ (-∞ -3] U (-1  0)

объединяем с первым

x∈ (-∞ - 3] U (-1  1) U (1 +∞)

Answer 2

To solve the inequality $\frac{1}{x-1}+\frac{3}{|x|+1}\geq\frac{1}{|x|-1}$, we need to consider two cases: one when $x$ is positive ($x>0$) and the other when $x$ is negative ($x<0$). This is because the absolute value function $|x|$ behaves differently for positive and negative values of $x$.

Case 1: $x>0$

In this case, the inequality becomes: $\frac{1}{x-1}+\frac{3}{x+1}\geq\frac{1}{x-1}$

Now, we need to find the domain for this inequality. The domain for $x>0$ is $x>0$.

Now, we can simplify the inequality:

$\frac{1}{x-1}+\frac{3}{x+1}-\frac{1}{x-1}\geq0$

$\frac{3}{x+1}\geq0$

Now, since $x>0$, the denominator $x+1$ is always positive, so we can multiply both sides of the inequality by $x+1$ without changing the inequality:

$3\geq0$

This inequality is always true, which means that for $x>0$, the original inequality holds true for all values of $x>0$.

Case 2: $x<0$

In this case, the inequality becomes: $\frac{1}{x-1}+\frac{3}{-x+1}\geq\frac{1}{-x-1}$

Now, we need to find the domain for this inequality. The domain for $x<0$ is $x<0$.

Now, we can simplify the inequality:

$\frac{1}{x-1}-\frac{3}{x-1}\geq\frac{1}{-x-1}$

$-\frac{2}{x-1}\geq\frac{1}{-x-1}$

Next, we need to consider two sub-cases:

Sub-case 2.1: $-1<x<1$

In this sub-case, the denominators $x-1$ and $-x-1$ are both negative, and thus we need to flip the inequality when multiplying or dividing by them:

$\frac{2}{x-1}\leq-\frac{1}{-x-1}$

$\frac{2}{x-1}\leq\frac{1}{x+1}$

Now, we need to find the domain where both sides of the inequality are positive. For $x<0$, $x+1$ is always positive, and for $-1<x<1$, $x-1$ is negative. Thus, this sub-case has no valid solutions.

Sub-case 2.2: $x<-1$ or $x>1$

In this sub-case, both $x-1$ and $-x-1$ are negative, so we don't need to flip the inequality:

$-\frac{2}{x-1}\geq-\frac{1}{-x-1}$

$-\frac{2}{x-1}\geq\frac{1}{x+1}$

Now, we need to find the domain where both sides of the inequality are negative. For $x<0$, $x-1$ is negative, and for $x<-1$ or $x>1$, $x+1$ is also negative. Therefore, this sub-case has valid solutions.

However, if we look at the original inequality, we notice that $x$ cannot be less than or equal to 0 0