Zad.2
a)
\frac{3}{x+2}+\frac{4}{x-3}=\frac{3(x-3)+4(x+2)}{(x+2)(x-3)}=\frac{3x-9+4x+8}{x^2-3x+2x-6}=\frac{7x-1}{x^2-x-6}
b)
\frac{2}{x-4}-\frac{3}{x-1}=\frac{2(x-1)-3(x-4)}{(x-4)(x-1)}=\frac{2x-2-(3x-12)}{x^2-x-4x+4}=\frac{2x-2-3x+12}{x^2-5x+4}=\frac{10-x}{x^2-5x+4}
c)
\frac{3x-6}{x-1}+\frac{6x-1}{2x+2}=\frac{3(x-2)*2(x+1)}{(x-1)*2(x+1)}=\frac{6(x^2+x-2x-2)}{2(x^2-1)}=\frac{3(x^2-x-2)}{x^2-1}
d)
\frac{3x-1}{x}-\frac{x-7}{2x-4}=\frac{(3x-1)(2x-4)-x(x-7)}{x(2x-4)}=\frac{6x^2-12x-2x+4-x^2+7x}{2x^2-4x}=
=\frac{5x^2-7x+4}{2x^2-4x}
e)
\frac{x}{x+5}+\frac{2-3x}{3x-1}=\frac{x(3x-1)+(2-3x)(x+5)}{(x+5)(3x-1)}=\frac{3x^2-x+2x+10-3x^2-15x}{3x^2-x+15x-5}=
=\frac{10-14x}{3x^2+14x-5}
f)
\frac{2x+1}{6-x}-\frac{3-2x}{x+6}=\frac{(2x+1)(x+6)-(3-2x)(6-x)}{(6-x)(6+x)}=\frac{2x^2+12x+x+6-(18-3x-12x+2x^2)}{6^2-x^2}=
=\frac{2x^2+13x+6-18+15x-2x^2}{36-x^2}=\frac{28x-12}{36-x^2}