Question: Solve: $\dfrac{x -1}{√x +1} = 1$

Solution:
Given,

$\dfrac{x -1}{√x +1} = 1$

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

$or, \dfrac{(√x +1)(√x -1)}{√x +1} = 1$

$or, √x -1 = 1$

$or, √x = 1 +1$

$or, √x = 2$

[ Squaring both sides ]

$or, (√x)^2 = 2^2$

$\therefore x = 4$
= Answer