Solve the given equation for $x$ and then determine the real number $m$ such that $|x| < 1$. The equation is: $\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{x^2+3x+2}$.

1. Problem Description

Solve the given equation for $x$ and then determine the real number $m$ such that $|x| < 1$. The equation is:

$\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{x^2+3x+2}$.

2. Solution Steps

First, we factor the quadratic expression in the denominator on the right-hand side:

$x^2+3x+2 = (x+1)(x+2)$.

Now, we rewrite the given equation as:

$\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{(x+1)(x+2)}$.

We multiply both sides of the equation by $(m-1)(x+1)(x+2)$ to clear the denominators. We assume $m \neq 1$, $x \neq -1$, and $x \neq -2$. Then

$(2m-5)(x+1) - 3(m-1)(x+2) = (3x+4)(m-1)$.

Expanding each term, we get:

$2mx + 2m - 5x - 5 - 3m(x+2) + 3(x+2) = 3mx + 4m - 3x - 4$

$2mx + 2m - 5x - 5 - 3mx - 6m + 3x + 6 = 3mx + 4m - 3x - 4$

$-mx - 4m - 2x + 1 = 3mx + 4m - 3x - 4$

Combine the terms with $x$:

$-mx - 3mx - 2x + 3x = 4m + 4 - 1 + 4m$

$-4mx + x = 8m + 3$

$x(1-4m) = 8m + 3$

If $1-4m \neq 0$, then

$x = \frac{8m+3}{1-4m}$.

We want to find $m$ such that $|x| < 1$. That is

$|\frac{8m+3}{1-4m}| < 1$

$-1 < \frac{8m+3}{1-4m} < 1$

We need to consider two cases: $1-4m > 0$ and $1-4m < 0$.

Case 1: $1-4m > 0 \Leftrightarrow 4m < 1 \Leftrightarrow m < \frac{1}{4}$.

Then, $-(1-4m) < 8m+3 < 1-4m$.

$-1+4m < 8m+3 \Leftrightarrow -4 < 4m \Leftrightarrow m > -1$.

$8m+3 < 1-4m \Leftrightarrow 12m < -2 \Leftrightarrow m < -\frac{1}{6}$.

So, in this case, $-1 < m < -\frac{1}{6}$.

Case 2: $1-4m < 0 \Leftrightarrow m > \frac{1}{4}$.

Then, $-(1-4m) > 8m+3 > 1-4m$.

$-1+4m > 8m+3 \Leftrightarrow -4 > 4m \Leftrightarrow m < -1$. This contradicts $m > \frac{1}{4}$, so there is no solution in this case.

$8m+3 > 1-4m \Leftrightarrow 12m > -2 \Leftrightarrow m > -\frac{1}{6}$.

Since we must have $m \neq 1$, and $m < \frac{1}{4}$ in Case 1, our solution becomes $-1 < m < -\frac{1}{6}$.

If $m=1$, then the original equation becomes

$\frac{-3}{x+2} - \frac{3}{x+1} = \frac{3x+4}{(x+1)(x+2)}$

$-3(x+1) - 3(x+2) = 3x+4$

$-3x-3-3x-6=3x+4$

$-6x-9 = 3x+4$

$-13 = 9x$

$x = -\frac{13}{9}$. Then $|x| = \frac{13}{9} > 1$, so $m=1$ is not a solution.

Finally, we check the condition $x\neq -1$ and $x\neq -2$.

If $x = -1$, then $\frac{8m+3}{1-4m} = -1$, so $8m+3 = -1+4m$, $4m = -4$, $m = -1$.

If $x = -2$, then $\frac{8m+3}{1-4m} = -2$, so $8m+3 = -2+8m$, $3=-2$, which is impossible.

Thus, we require $m \neq -1$. Since the interval of solutions is $-1 < m < -\frac{1}{6}$, $m$ cannot be equal to $-1$, so the solution is $-1 < m < -\frac{1}{6}$.

3. Final Answer

$-1 < m < -\frac{1}{6}$