Question

ME AJUDEM, É URGENTE 
Simplifique as frações admitindo que os denominadores sejam diferentes de zeroa) (x² - 4) / (x – 2) =
b) (a² - 9) / 5(a + 3) =
c) (4x² - y²) / ( 2x – y) =
d) (a + b)⁵ / (a + b)² =
e) ( a – b)² / ( a² - b²) =
f) (x + y)² / ( x² - y²) =
g) (x² - 2x + 1) / (x² - 1) =
h) ( a + 1) / (a² + 2 a + 1) =
i) (x² + 6x + 9) / (2x + 6) =

Answer 1

x² - 4) / (x – 2) =
b) (a² - 9) / 5(a + 3) =  (a-3) (a+3) /  5(a+3) = (a-3)/5
c) (4x² - y²) / ( 2x – y) = [(2x+y)(2x-y)]/ (2x-y)  = (2x + y)
d) (a + b)⁵ / (a + b)² = (a + b)5-2  =   (a + b)3
e) ( a – b)² / ( a² - b²) =  ( a – b)( a – b) / (a+b)(a-b)  = ( a – b) / (a+b)
f) (x + y)² / ( x² - y²) = (x + y)(x + y) / ( x + y) ( x - y)  = (x + y) / ( x - y)
g) (x² - 2x + 1) / (x² - 1) = (x-1)² /(x-1)(x+1)  =  (x-1) /(x+1) 
h) ( a + 1) / (a² + 2 a + 1) = ( a + 1) / (a + 1) ²  = 1/  ( a + 1)
i) (x² + 6x + 9) / (2x + 6) = (x + 3)² / 2(x + 3) =  (x + 3)/2