a,

`(3x)/(x+5)+(3x)/(x-5)` = `3x*(1/(x+5)+1/(x-5))` = `3x*(x-5+x+5)/(x^2-25)` = `(6x^2)/(x^2-25)`

b,

`(2c^2+2cx)/(x^3+c^3)-(c-x)/(x^2-xc+c^2)` = `(2c*cancel((c+x)))/(cancel((x+c))(x^2-xc+c^2))-(c-x)/(x^2-xc+c^2)` =

= `(2c)/(x^2-xc+c^2)-(c-x)/(x^2-xc+c^2)` = `(2c-(c-x))/(x^2-xc+c^2)` = `(c+x)/(x^2-xc+c^2)`

c, Itt a két tört közötti műveletet csak tippelni tudom, ha a négy feladat a négy alapművelet, akkor szorzás van.

`(12xy^4)/(5a^4b)*(5a^3b)/(4xy^4)` = `(12x*y^4*5a^3b)/(5a^4b*4xy^4)` = `60/(20a)` = `3/a`

d,

`(x^2+4x+4)/(x-3) : (x+2)/(x^2-9)` = `(x+2)^(cancel(2))/cancel((x-3)) : cancel((x+2))/((x+3)*cancel((x-3)))` = `(x+2):1/(x+3)` = `(x+2)(x+3)` = `x^2+5x+6`