((5)/(x-1)) - ((2x)/(x+1)) < 1
( 5(x+1)-2x(x-1) ) / (x^2 - 1) < 1
( 5x + 5 -2x^2 + 2x ) / (x^2 - 1) < 1
( -2x^2 + 7x + 5 ) / (x^2 - 1) < 1
-2x^2 + 7x + 5 = 0
x = (-7 +- sqrt(7^2 - 4(-2)(5)) / 2(-2)
x = (-7 +- sqrt(49 + 40)) / -4
x = (7 +- sqrt(89)) / 4
x = (7-sqrt(89))/4, (7+sqrt(89))/4
x = about -0.6805, 4.1085
x^2 - 1 = 0
x = -1, 1
We have four places where things can change: x = -1, -0.6805, 1, 4.1085
Let's test some values: x = -10, -0.7, 0, 2, 5
f(x) = ((5)/(x-1)) - ((2x)/(x+1))
f(-10) = 5/-11 - (-20/-9) = -5/11 - 20/9 = negative = less than 1
f(-0.7) = 5/-1.7 - (-1.4/0.3) = -5/1.7 + 1.4/0.3 = about 1.7255 = greater than 1
f(0) = 5/-1 - 0 = -5 = less than 1
f(2) = 5/1 - 4/3 = 15/3 - 4/3 = 11/3 = greater than 1
f(5) = 5/4 - 10/6 = 15/12 - 20/12 = -5/12 = less than 1
So we don't want the regions including x = -0.7 and 2, and we do want the regions including x = -10, 0, and 5.
Answer: (x can be any of these)
x < -1
(7-sqrt(89))/4 < x < 1
(7+sqrt(89))/4 < x