Question

Uprość wyrażenia:
g) (x^2-1)/|x+1| - |x-1|
h) (x^2-4)/|x|-2 + |x-2|
i) |x^4-1|/|x^2+1| * |1/(|x|-1)(|x|+1)|

Z góry dziękuję za pomoc :)

Answer 1

g)(x²-1)/|x+1| - |x-1|= ?
x+1 ≠ 0
x ≠-1

x > 1
            +        +
(x²-1)/|x+1| - |x-1| = (x²-1)/(x+1) - (x-1) =
= (x-1)(x+1)/(x+1) - x+1 = x-1 - x+1= 0

0 < x ≤ 1
            +         -
(x²-1)/|x+1| - |x-1| = (x²-1)/(x+1) - (1-x) =
= (x-1)(x+1)/(x+1) - 1+x = x-1 - 1+x = 2x-2

x <-1
             -         -
(x²-1)/|x+1| - |x-1| = (1-x²)/(x+1) - (1-x) =
(1-x)(x+1)/(x+1) - 1+x = 1-x - 1+x = 0

h)(x²-4)/(|x|-2) + |x-2|= ?
x ≠±2

x > 2
           +            +
(x²-4)/(|x|-2) + |x-2| = (x+2)(x-2)/(x-2) + x-2 =
= x+2 + x-2 = 2x

0 < x < 2
            +           -       
(x²-4)/(|x|-2) + |x-2| = (x+2)(x-2)/(x-2) + (2-x) =
= x+2 + 2-x = 2

x < 0
            -            -
(x²-4)/(|x|-2) + |x-2| = (x²-4)/(-x-2) + (2-x) =
= -(x-2)(x+2)/(x+2) + 2-x = 2-x + 2-x = 2(2-x)

i)|x⁴-1|/|x²+1| * |1/(|x|-1)(|x|+1)|= ?
x ≠±1

|x⁴-1|/|x²+1| * |1/(|x|-1)(|x|+1)| = |x⁴-1|/|x²+1| * |1/(|x|²-1)| =
= |x⁴-1|/|x²+1| * 1/|x²-1| = |x⁴-1|/|x²+1||x²-1| = |x⁴-1|/|x⁴-1| = 1