Al gebr oal mu'kabla

$\begin{aligned} \frac {x+2a}{x-2a} + \frac {x+2b}{x-2b} & = \frac {1+\frac {2a}x}{1-\frac {2a}x} + \frac {1+\frac {2b}x}{1-\frac {2b}x} \\ & = \frac {1+\frac a2\left(\frac 1a + \frac 1b\right)}{1-\frac a2\left(\frac 1a + \frac 1b\right)} + \frac {1+\frac b2\left(\frac 1a + \frac 1b\right)}{1-\frac b2\left(\frac 1a + \frac 1b\right)} \\ & = \frac {\frac 32 + \frac a{2b}}{\frac 12 - \frac a{2b}} + \frac {\frac 32 + \frac b{2a}}{\frac 12 - \frac b{2a}} \\ & = \frac {3b+a}{b-a} + \frac {3a+b}{a-b} \\ & = 3 + \frac {4a}{b-a} + 3 + \frac {4b}{a-b} \\ & = 6 - \frac {4a-4b}{a-b} \\ & = 6 - 4 \\ & = \boxed 2 \end{aligned}$

$a=\frac{xb}{4b-x}$

$x+2a = x + (\frac{2xb}{4b-x})$

$x-2a = x - (\frac{2xb}{4b-x})$

Multiplying each of the above by $4b-x$ …

$\frac{x+2a}{x-2a}=\frac{4bx-x^2+2bx}{4bx - x^2 -2bx} = \frac{6bx - x^2}{2bx - x^2}$

Now that we have $\frac{x+2a}{x-2a}$ in terms of $b$ , we can simplify the equation by adding $\frac{x+2b}{x-2b}$ as it is, without substitution.

Multiplying the top and bottom of $\frac{x+2b}{x-2b}$ by $-x$ gives each of the fractions the same denominator, so the result is:

$\frac{6bx - x^2 + (-2bx - x^2)}{2bx - x^2 + (2bx - x^2)}$

This simplifies to:

$\frac{4bx - 2x^2}{2bx-x^2}$ , or $2\frac{2bx - x^2}{2bx-x^2}$

Cancelling out the like terms just leaves, 2, so the answer is 2 .

Solving the first equation we get $x=\dfrac{4ab}{a+b}$ . So $\dfrac{x}{2a}=\dfrac{2b}{a+b}$ and $\dfrac{x}{2b}=\dfrac{2a}{a+b}$ . Hence $\dfrac{x+2a}{x-2a}=\dfrac{a+3b}{b-a}$ and $\dfrac{x+2b}{x-2b}=\dfrac{3a+b}{a-b}$ . Therefore the required sum is $\dfrac{2(a-b)}{a-b}=\boxed {2}$