Odpowiedź:
[tex]a)\\\\\frac{(x-4)(2x+1)}{1-x}=0\\\\1-x\neq 0\\\\-x\neq -1\ \ /\cdot(-1)\\\\x\neq 1\\\\(x-4)(2x+1)=0\\\\x-4=0\ \ \ \ \vee\ \ \ \ 2x+1=0\\\\x=4\ \ \ \ \ \ \ \ \vee\ \ \ \ 2x=-1\ \ /:2\\\\x=4\ \ \ \ \ \ \ \ \vee\ \ \ \ 2x=-\frac{1}{2}[/tex]
[tex]b)\\\\\frac{x+1}{x+2}-\frac{x-2}{x-3}=0\\\\x+2\neq 0\ \ \ \ i\ \ \ \ x-3\neq 0\\\\x\neq -2\ \ \ \ \ \ i\ \ \ \ x\neq 3\\\\\frac{(x-3)(x+1)}{(x+2)(x-3)}-\frac{(x+2)(x+2)}{(x+2)(x-3)}=0\\\\\frac{(x-3)(x+1)-(x+2)^2}{(x+2)(x-3)}=0\\\\\frac{x^2+x-3x-3-(x^2+4x+4)}{(x+2)(x-3)}=0\\\\\frac{x^2-2x-3-x^2-4x-4}{(x+2)(x-3)}=0\\\\\frac{-6x-7}{(x+2)(x-3)}=0\\\\-6x-7=0\\\\-6x=7\ \ /:(-6)\\\\x=-\frac{7}{6}\\\\x=-1\frac{1}{6}[/tex]
