The problem asks to solve the given equations for the unknown variable $x$ and discuss the solution based on the parameters $a, b, c, m, n$. We will solve the first two equations. 1) $a(ax+1) = 2(2x-1)$ 2) $\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2-b^2}$

1. Problem Description

The problem asks to solve the given equations for the unknown variable $x$ and discuss the solution based on the parameters $a, b, c, m, n$. We will solve the first two equations.

1) $a(ax+1) = 2(2x-1)$

2) $\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2-b^2}$

2. Solution Steps

1) $a(ax+1) = 2(2x-1)$

$a^2x + a = 4x - 2$

$a^2x - 4x = -a - 2$

$x(a^2-4) = -(a+2)$

$x(a-2)(a+2) = -(a+2)$

If $a \neq 2$ and $a \neq -2$, then

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

If $a = -2$, then

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

$0 = 0$

So $x$ can be any real number.

If $a = 2$, then

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

$0 = -4$

This is not possible, so there is no solution.

2) $\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2-b^2}$

$\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{(a-b)(a+b)}$

Multiply both sides by $(b-a)(a+b)$:

$(a-x)(a+b) - (a+x)(b-a) = \frac{2ax(b-a)(a+b)}{(a-b)(a+b)} = \frac{2ax(b-a)}{(a-b)} = -2ax$

$a^2+ab-ax-bx - (ab -a^2 + bx - ax) = -2ax$

$a^2+ab-ax-bx - ab + a^2 - bx + ax = -2ax$

$2a^2 - 2bx = -2ax$

$2a^2 = 2bx - 2ax$

$a^2 = bx - ax$

$a^2 = x(b-a)$

If $b \neq a$, then

$x = \frac{a^2}{b-a}$

If $b = a$, then

$a^2 = x(a-a)$

$a^2 = 0$

$a=0$

If $a=0$ and $b=0$, any $x$ is a solution as the original equation reduces to $\frac{-x}{0} - \frac{x}{0} = 0$. However, since division by zero is undefined, there is no solution.

3. Final Answer

1) If $a \neq 2$ and $a \neq -2$, $x = \frac{1}{2-a}$.

If $a = -2$, $x$ can be any real number.

If $a = 2$, there is no solution.

2) If $b \neq a$, $x = \frac{a^2}{b-a}$.

If $b = a$, then the equation is undefined, and there is no solution.