Question

If $x=\frac{6ab}{a+b},$ find the value of :
$\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}.$

Answer 1

$x=\frac{6ab}{a+b}$

$\Rightarrow \frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}$

$\Rightarrow \frac{(x+3a)(x-3b)+(x-3a)(x+3b)}{(x-3a)(x-3b)}$

$\Rightarrow \frac{x^2-3bx+3ax-9ab+x^2-3ax+3bx-9ab}{(x-3a)(x-3b)}$

$\Rightarrow \frac{2x^2-18ab}{(x-3a)(x-3b)}$

$\Rightarrow \frac{2{\left(\frac{6ab}{a+b}\right)}^2-18ab}{(\frac{6ab}{a+b}-3a)(\frac{6ab}{a+b}-3b)}$

$\Rightarrow \frac{2(6ab{)}^2-18ab(a+b{)}^2}{[6ab-3a(a+b\left)\right][6ab-3b(a+b\left)\right]}$

$\Rightarrow \frac{2\left[\right(6ab{)}^2-9ab(a+b{)}^2]}{(6ab{)}^2-9ab(a+b{)}^2}$

$\Rightarrow 2$.