Question

solve by componento or divedendo

Answer 1

we can write following questionlike this
(1-ax/1+ax)^2=1-bx/1+bx
then apply com. and div. rule both side

Answer 2

We know that,

if $\frac{a}{b}$ = $\frac{c}{d}$ then,

$\frac{a+b}{a-b}$ = $\frac{c+d}{c-d}$.

The given question is: ($\frac{1-ax}{1+ax}$)$\sqrt{\frac{1+bx}{1-bx}$ = 1.

or, squaring both side we get,

($\frac{1-ax}{1+ax}$)² = ($\frac{1-bx}{1+bx}$)

or, applying C & D we get,

($\frac{1-2ax+a²x²+1+2ax+a²x²}{1-2ax+a²x²-1-2ax-a²x²}$) = $\frac{1-bx+1+bx}{1-bx-1-bx}$

or, $\frac{2(1+a²x²)}{2(-2ax)}$ = $\frac{2}{-2bx}$

or, 2bx + 2a²bx³ = 4ax.

or, 2x(b+a²bx²) = 2x(2a)

or, a²bx² + b = 2a

or, a²bx² = 2a - b

or, x² = (2a - b)/a²b

or, x = √[(2a - b)/a²b] [Ans]