e)
\frac{(x-1)^2+4x}{x^2-1}=\frac{x^2-2x+1+4x}{x^2-1}=\frac{x^2+2x+1}{x^2-1}=\frac{(x+1)^2}{(x-1)(x+1)}=\frac{(x+1)(x+1)}{(x-1)(x+1)}=\frac{x+1}{x-1}
x^2-1\ne0 => x^2\ne1 => x\ne -1 , x\ne 1
f)
\frac{(x+4)^2-16x}{(x-4)^2}=\frac{x^2+8x+16-16x}{(x-4)^2}=\frac{x^2-8x+16}{(x-4)^2}=\frac{(x-4)^2}{(x-4)^2}=1
x\ne 4
g)pracadomowa24.pl skracanie i rozszerzanie ułamków zadania i rozwiązania
\frac{4-(x-2)^2}{16x-8x^2+x^3}=\frac{4-(x^2-4x+4)}{x(x^2-8x+16)}=\frac{4-x^2+4x-4}{x(x-4)^2}=
x\ne 0 , x\ne 4
\frac{-x^2+4x}{x(x-4)^2}=\frac{-x(x-4)}{x(x-4)(x-4)}=\frac{-1}{x-4}=\frac{-1}{-(4-x)}=\frac{1}{4-x}
h)
\frac{x^3+10x^2+25x}{[1-(x+4)^2]x}=\frac{x(x^2+10x+25)}{[1-(x^2+8x+16)]x}=\frac{(x+5)^2}{1-x^2-8x-16}=
x\ne 0 , x\ne -3, x\ne -5
=\frac{(x+5)^2}{-x^2-8x-15}=\frac{(x+5)^2}{-(x^2+3x+5x+15)}=\frac{(x+5)^2}{-[x(x+3)+5(x+3)]}=
\frac{(x+5)(x+5)}{-(x+3)(x+5)}=-\frac{x+5}{x+3}