a) (x-1)/3(x-4) - (x-3)/2(x-4)
Conditia de existenta:
x-4≠0⇒x≠4
(2x-2-3x+9)/6(x-4)=(-x+7)/6(x-4)
b) 2y/2(y-2) - 3y/3(y-2)=
y≠2
La prima fractie cred ca ai redactat gresit si nu este inmultire ci scadere.
y/(y-2) - y/(y-2)=0
c) x/(2x-6) - (x+1)/(3-x)=
x/2(x-3) + (x+1)/(x-3)=
(x+2x+2)/2(x-3)=
(3x+2)/2(x-3)
x≠3