Odpowiedź:
a)[tex]\frac{4x-12}{x+2} *\frac{2x+4}{2x-6)}=\frac{4*(x-3)}{x+2} *\frac{2(x+2)}{2(x-3)} =4[/tex]
x+2≠ 0 i 2x-6≠0
x≠-2 2x≠6 ⇔ x≠3 D:x∈R-{-2;3}
b)[tex]\frac{3-x}{x} *\frac{2x+6}{x^2-9} =\frac{-(x-3)}{x} *\frac{2(x+3)}{(x-3)(x+3)} =\frac{-2}{x}[/tex]
zał:x≠0 i x²-9≠0
x²≠9⇔⇔ x≠-3 i x≠3 D:x∈R-{-3;0;3}
c)[tex]\frac{-6}{3-2x} :\frac{9}{4x-6} =\frac{-6}{3-2x} *\frac{2(2x-3)}{9} =\frac{-6}{-(2x-3)} *\frac{2(2x-3)}{9} =\frac{2}{3}[/tex]
zał 3-2x≠0 i 4x-6≠0
-2x≠-3 4x≠6
x≠3/2 x≠3/2 D:x∈R-{3/2}
d)[tex]\frac{16}{3x+1} :\frac{12}{9x+3} =\frac{16}{3x+1} *\frac{3(3x+1)}{12} =4[/tex]
zał 3x+1≠0 i 9x+3≠0
3x≠-1 i 9x≠-3
x≠-1/3 i x≠-21/3 D:x∈R-{-1/3}
e)[tex]\frac{x}{x+5} +\frac{2-3x}{3x-1} =\frac{x(3x-1)}{(x+5)(3x-1)} +\frac{(2-3x)(x+5)}{(3x-1)(x+5)} =\frac{3x^2-x+2x+10-3x^2-15x}{(x+5)(3x-1)} =\frac{-14x+10}{(x+5)(3x-1)}[/tex]
zał: x+5≠0 i 3x-1≠0
x≠-5 3x≠1 ⇔ x≠1/3 x∈R-{-5; 1/3}
f)[tex]\frac{2x+1}{6-x} -\frac{3-2x}{x+6} =\frac{(2x+1)(6+x)}{(6-x)(6+x)} -\frac{(3-2x)(6-x)}{(6+x)(6-x)} =\frac{12x+2x^2+6+x-(18-3x-12x+2x^2)}{x^2-36} =\frac{2x^2+13x+6-2x^2+15x-18}{x^2-36} =\frac{28x-12}{x^2-36}[/tex]
zał:6-x≠0 x+6≠0
x≠0 x≠-6 D:x∈-R{-6;6}
Szczegółowe wyjaśnienie:
