Question: Solve: $\dfrac{x-1}{\sqrt{x} +1 } = 4 + \dfrac{\sqrt{x} -1}{2}$

Solution:
Given,

$\dfrac{x-1}{\sqrt{x} +1 } = 4 + \dfrac{\sqrt{x} -1}{2}$

$or, \dfrac{(\sqrt{x})^2 - (1)^2}{\sqrt{x} +1 } =  4+ \dfrac{\sqrt{x} -1}{2}$

$or, \dfrac{(\sqrt{x} +1 )(\sqrt{x} -1)}{(\sqrt{x) +1)} = 4 + \dfrac{\sqrt{x} -1}{2}$

$or, \sqrt{x} - 1 = 4 + \dfrac{\sqrt{x} -1}{2}$

$or, (\sqrt{x} - 1 -  \dfrac{\sqrt{x} -1}{2} = 4$

$or, \dfrac{2(\sqrt{x}-1) - (\sqrt{x} -1}{2} = 4$

$or, 2\sqrt{x} - 2 - \sqrt{x} + 1 = 4*2$

$or, \sqrt{x} - 1 = 8$

$or, \sqrt{x} = 8 +1$

$or, \sqrt{x} = 9$

squaring both sides

$or, (\sqrt{x})^2 = 9^2$

$\therefore x = 81$