Expand and simplify the expressions: (x-y)2, (4x-2y)2. Solve for x: x2+5x+6=0, 6/(x−4) = (x+4)/x

Here, $a=x$ and $b=y$.

$$(x-y)^2 = x^2 - 2(x)(y) + y^2$$

$$x^2 - 2xy + y^2$$

The expanded and simplified expression is $\boxed{x^2 - 2xy + y^2}$.

b) $(4x-2y)^2$

Here, $a=4x$ and $b=2y$.

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

$$(4x)^2 = 16x^2$$ $$2(4x)(2y) = 16xy$$ $$(2y)^2 = 4y^2$$

$$16x^2 - 16xy + 4y^2$$

The expanded and simplified expression is $\boxed{16x^2 - 16xy + 4y^2}$.

2. Solve for $x$:

a) $x^2+5x+6=0$

Find two numbers that multiply to 6 and add to 5. These numbers are 2 and 3.

$$(x+2)(x+3) = 0$$

$$x+2 = 0 \quad \text{or} \quad x+3 = 0$$ $$x = -2 \quad \text{or} \quad x = -3$$

The solutions are $\boxed{x = -2 \text{ or } x = -3}$.

b) $\frac{6}{x-4} = \frac{x+4}{x}$

Note that $x \neq 4$ and $x \neq 0$.

$$6x = (x+4)(x-4)$$

$$6x = x^2 - 4x + 4x - 16$$ $$6x = x^2 - 16$$

Find two numbers that multiply to -16 and add to -6. These numbers are -8 and 2.

$$(x-8)(x+2) = 0$$

$$x-8 = 0 \quad \text{or} \quad x+2 = 0$$ $$x = 8 \quad \text{or} \quad x = -2$$

Both solutions are valid as they do not make the original denominators zero.
The solutions are $\boxed{x = 8 \text{ or } x = -2}$.