Question

If \(x=\dfrac{4ab}{a+b}\), evaluate \(\displaystyle \frac{x+2a}{x-2a}+\frac{x+2b}{x-2b}\).

Hint:

Before substituting values, simplify the algebraic expression—often large cancellations become visible and the substitution becomes trivial.

Answer 1

Combine over a common denominator: \[ \frac{x+2a}{x-2a}+\frac{x+2b}{x-2b} =\frac{(x+2a)(x-2b)+(x+2b)(x-2a)}{(x-2a)(x-2b)} =\frac{2x^2-8ab}{x^2-2(a+b)x+4ab}. \] Now substitute \(x=\dfrac{4ab}{a+b}\). Then \[ \text{Numerator} = 2\!\left(x^2-4ab\right) = -\,\frac{8ab\,(a-b)^2}{(a+b)^2}, \] \[ \text{Denominator} = x^2-2(a+b)x+4ab = -\,\frac{4ab\,(a-b)^2}{(a+b)^2}. \] Hence the value is \[ \frac{-\,\frac{8ab\,(a-b)^2}{(a+b)^2}}{-\,\frac{4ab\,(a-b)^2}{(a+b)^2}}=2. \] \[ \boxed{2} \]