Question

A+√(a2-2ax)/a-√(a2-2ax)=b find x

Answer 1

$\textbf{Given:}$

$\dfrac{a+\sqrt{a^2-2ax}}{a-\sqrt{a^2-2ax}}=b$

$\textbf{Given:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{a+\sqrt{a^2-2ax}}{a-\sqrt{a^2-2ax}}=b$

$\implies\,a+\sqrt{a^2-2ax}=b(a-\sqrt{a^2-2ax})$

$\implies\,a+\sqrt{a^2-2ax}=ab-b\sqrt{a^2-2ax}$

$\implies\,\sqrt{a^2-2ax}+b\sqrt{a^2-2ax}=ab-a$

$\implies\,\sqrt{a^2-2ax}(1+b)=ab-a$

$\implies\,\sqrt{a^2-2ax}=\dfrac{ab-a}{1+b}$

$\text{Squaring on bothsides, we get}$

$a^2-2ax=\dfrac{(ab-a)^2}{(1+b)^2}$

$a^2-2ax=\dfrac{a^2b^2+a^2-2a^2b}{1+b^2+2b}$

$a^2-\dfrac{a^2b^2+a^2-2a^2b}{1+b^2+2b}=2ax$

$\dfrac{a^2(1+b^2+2b)-(a^2b^2+a^2-2a^2b)}{1+b^2+2b}=2ax$

$\dfrac{a^2+a^2b^2+2a^2b-a^2b^2-a^2+2a^2b}{1+b^2+2b}=2ax$

$\dfrac{2a^2b+2a^2b}{1+b^2+2b}=2ax$

$\dfrac{4a^2b}{1+b^2+2b}=2ax$

$\implies\,x=\dfrac{2ab}{1+b^2+2b}$

$\implies\bf\,x=\dfrac{2ab}{(1+b)^2}$

$\therefore\textbf{The value of x is \bf\,\dfrac{2ab}{(1+b)^2} }$

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Answer 2

hope it will help you all