Я разложила по действиям

д)

1. (1+с)(4с^2+2c+1)/(2c-1)(4c^2+2c+1)(1+2c)=(1+c)/(2c-1)(2c+1)

2. 1/2(1-2c) + (1+c)/(2c-1)(2c+1) = (1+2c-2-2c)/2(1-2c)(1+2c) = -1/2(1-2c)(1+2c)

3.-2(2c-1)/2(1-2c)(1+2c)^2 = 1/(1+2c)^2

4. 1/(1+2c)^2 - 1/(1+2c)^2 = 0

e)

 1. (b+1)(4b^2+2b+1)/(2b-1)(4b^2+2b+1)(1+2b) = (b+1)/(2b-1)(1+2b)

 2. 1/2(1-2b)+(b+1)/(2b-1)(1+2b) = (1+2b-2b-2)/2(1-2b)(1+2b) = -1/2(1-2b)(1+2b)

 3. -2(2b-1)/2(1-2b)(1+2b)^2 = 1/(2b+1)^2

 4. 1/(2b+1)^2 - 1/(2b+1)^2 = 0