I tried to solve thees but is useless I not doing it right. Can some one lead the way?
1) sqrt(2x + 1) = x-1
2) -2< x+2/x-1
$\displaystyle \sqrt{2x+1} = x-1$
square both sides ...
$\displaystyle 2x+1 = x^2 - 2x + 1$
$\displaystyle 0 = x^2 - 4x $
$\displaystyle 0 = x(x-4)$
$\displaystyle x = 0 , x = 4$
check for extraneous solution(s)
$\displaystyle -2 < \frac{x+2}{x-1}$
$\displaystyle 0 < 2 + \frac{x+2}{x-1}$
$\displaystyle 0 < \frac{2(x-1)}{x-1} + \frac{x+2}{x-1}$
$\displaystyle 0 < \frac{3x}{x-1}$
critical values are $\displaystyle x = 0$ and $\displaystyle x = 1$. these two values of x break up the set of all x-values into three sections . ..
$\displaystyle x < 0$ , $\displaystyle 0 < x < 1$ , and $\displaystyle x > 1$
pick any number in each section ... if it makes the original inequality true, then all values of x in that section will be in the solution set.