d)
\frac{x+1}{x+5}-\frac{x+3}{x+2}+x=\frac{(x+1)(x+2)}{(x+5)(x+2)}-\frac{(x+3)(x+5)}{(x+5)(x+2)}+\frac{x(x+5)(x+2)}{ (x+5)(x+2) }=
=\frac{x^2+2x+x+2-(x^2+5x+3x+15)+x(x^2+2x+5x+10)}{x^2+2x+5x+10}=
=\frac{x^2+3x+2-x^2-8x-15+x^3+7x^2+10x}{ x^2+7x+10 }=\frac{x^3+7x^2+5x-13}{x^2+7x+10 }
f)
\frac{3x-1}{2x+4}-\frac{2-x}{ x+2 }-\frac{1}{ 2x-4 }=\frac{(3x-1)( x+2)(2x-4)}{ (4x^2-16)(x+2) }-\frac{(2-x)(4x^2-16)}{ (4x^2-16) (x+2) }-\frac{(x+2)(2x+4)}{ (4x^2-16) (x+2) }=
=\frac{(3x-1)(2x^2-4x+4x-8)-(8x^2-32-4x^3+16x)-(2x^2+4x+4x+8)}{ 4x^3+8x^2-16x-32 }=
=\frac{6x^3-24x-2x^2+8+4x^3-8x^2-16x+32-2x^2-8x-8}{4x^3+8x^2-16x-32 }=
=\frac{10x^3-12x^2-48x+32}{4(x^3+2x^2-4x-8) }=\frac{2(5x^3-6x^2-24x+16)}{4(x^3+2x^2-4x-8) }=
=\frac{5x^3-6x^2-24x+16}{2(x^3+2x^2-4x-8) }=\frac{5x^2(x+2)-16x(x+2)+8(x+2)}{2((x^2(x+2)-4(x+2)) }=\frac{(x+2)(5x^2-16x+8)}{2(x+2)(x^2+4)}=
=\frac{5x^2-16x+8}{2(x^2+4)}