We are asked to solve the following equations for $x$, where $a$ is a real parameter: a) $1 - 3a + 5x = ax - 2a + 4x$ b) $(x + a)^2 = x(x + a) + 4a^2$ c) $a(x - \frac{1}{a^2}) + a^2(x - \frac{1}{a}) = 2$

1. Problem Description

We are asked to solve the following equations for $x$, where $a$ is a real parameter:

a) $1 - 3a + 5x = ax - 2a + 4x$

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

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

2. Solution Steps

a) $1 - 3a + 5x = ax - 2a + 4x$

Rearrange the equation to isolate terms with $x$ on one side and constant terms on the other side:

$5x - ax - 4x = -2a + 3a - 1$

$x - ax = a - 1$

$x(1 - a) = a - 1$

If $a \ne 1$, we have $x = \frac{a - 1}{1 - a} = \frac{-(1 - a)}{1 - a} = -1$.

If $a = 1$, we have $x(1 - 1) = 1 - 1$, which gives $0 = 0$. This means that the equation is true for all real values of $x$.

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

Expand the terms:

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

$2ax + a^2 = ax + 4a^2$

$2ax - ax = 4a^2 - a^2$

$ax = 3a^2$

If $a \ne 0$, then $x = \frac{3a^2}{a} = 3a$.

If $a = 0$, then $0 \cdot x = 3 \cdot 0^2$, which gives $0 = 0$. This means that the equation is true for all real values of $x$.

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

Distribute $a$ and $a^2$:

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

$ax - \frac{1}{a} + a^2x - a = 2$

$ax + a^2x = 2 + a + \frac{1}{a}$

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

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

If $a(1 + a) \ne 0$, then $x = \frac{(a + 1)^2}{a} \cdot \frac{1}{a(a + 1)} = \frac{a + 1}{a^2}$.

If $a = 0$, the original equation is not defined, so $a$ cannot be

0. If $a = -1$, then $x(-1(1 - 1)) = \frac{(-1 + 1)^2}{-1}$, so $0 = 0$, which means the equation is true for all $x$. However, we must consider the initial equation, $a(x - \frac{1}{a^2}) + a^2(x - \frac{1}{a}) = 2$. Since $\frac{1}{a}$ and $\frac{1}{a^2}$ exist, $a \ne 0$. So when $a = -1$, $-1(x - \frac{1}{1}) + 1(x - \frac{1}{-1}) = 2$ gives $-x + 1 + x + 1 = 2$, which is $2 = 2$. Thus, when $a = -1$, the equation is true for all $x$.

3. Final Answer

a) If $a \ne 1$, $x = -1$. If $a = 1$, the equation is true for all $x$.

b) If $a \ne 0$, $x = 3a$. If $a = 0$, the equation is true for all $x$.

c) If $a \ne 0$ and $a \ne -1$, $x = \frac{a + 1}{a^2}$. If $a = -1$, the equation is true for all $x$.