Zadanie 9
c)
\frac{x^2}{3-x}*\frac{3x-9}{x^4}=\frac{x^2}{3-x}*\frac{-3(3-x)}{x^4}=\frac{-3x^2}{x^4}=-\frac{3}{x^2}
D=\mathbb R\backslash \{0,3\}
Zadanie 10
a)
{\frac{x-1}{x+1}*\frac{x^3+x^2}{x^4-1}=\frac{x-1}{x+1}*\frac{x^2(x+1)}{(x^2-1)(x^2+1)}=\frac{x-1}{1}*\frac{x^2(x+1)}{(x-1)(x+1)(x^2+1)}=\frac{x^2}{(x-1)(x^2+1)}}
dziedzina:
x+1\ne0
x\ne -1
i
x^4-1\ne0
x^4\ne1
x\ne -1 i x\ne 1
D=\mathbbR \backslash \{-1,1\}
c)
\frac{x^2-4}{x^3+3x^2}*\frac{x^4-9x^2}{x^2+2x}=\frac{(x-2)(x+2)}{x^2(x+3)}*\frac{x^2(x^2-9)}{x(x+2)}=\frac{x-2}{x+3}*\frac{(x-3)(x+3)}{x}=
=\frac{(x-2)(x-3)}{x}=\frac{x^2-3x-2x+6}{x}=\frac{x^2-5x+6}{x}
dziedzina:
x^3+3x^2\ne0
x^2(x+3)\ne0
x\ne 0 , x\ne -3
i
x^2+2x\ne0
x(x+2)\ne0
x\ne 0 , x\ne -2
D=\mathbbR\backslash \{-3,-2,0\}
d)
\frac{x^2-16}{x+1}*\frac{1-x^2}{x+4}=\frac{(x-4)(x+4)}{x+1}*\frac{-(x^2-1)}{x+4}=\frac{x-4}{x+1}*\frac{-(x-1)(x+1)}{1}=
=-(x-4)(x-1)=-(x^2-x-4x+4)=-x^2+5x-4
dziedzina
x\ne -1 i x\ne -4
D=\mathbbR \backslash \{-4,-1\}