Question

Wykonaj działania, podaj dziedzinę wyrażenia. Przedstaw wynik w najprostszej postaci.

Answer 1

Odpowiedź:

a)

[tex]x\in R\setminus \lbrace0,1\rbrace\\\frac{1}{x}+\frac{2+x}{x-1}=\frac{x-1}{x(x-1)}+\frac{x(2+x)}{x(x-1)}=\frac{x-1+x(2+x)}{x(x-1)}=\frac{x-1+2x+x^2}{x(x-1)}=\frac{x^2+3x-1}{x^2-x}[/tex]

b)

[tex]x\in R\setminus\lbrace-3,0,3\rbrace\\\frac{x^2+4}{x^2-3x}-\frac{1}{x^2-9}=\frac{x^2+4}{x(x-3)}-\frac{1}{(x-3)(x+3)}=\frac{(x^2+4)(x+3)}{x(x-3)(x+3)}-\frac{x}{x(x-3)(x+3)}=\frac{(x^2+4)(x+3)-x}{x(x-3)(x+3)}=\frac{x^3+3x^2+4x+12-x}{x(x-3)(x+3)}=\frac{x^3+3x^2+3x+12}{x(x^2-9)}=\frac{x^3+3x^2+3x+12}{x^3-9x}[/tex]

c)

[tex]x\in R \setminus \lbrace-1,1,3\rbrace\\\frac{2x}{x^2-1}*\frac{x+1}{2x-6}=\frac{2x}{(x+1)(x-1)}*\frac{x+1}{2(x-3)\\}=\frac{x}{x-1}*\frac{1}{x-3}=\frac{x}{(x-1)(x-3)}=\frac{x}{x^2-4x+3}[/tex]

d)

[tex]x\in R \setminus \lbrace-\frac{1}{2},0,3\rbrace\\\frac{8x-24}{2x^3+x^2}:\frac{x-3}{2x+1}=\frac{8x-24}{2x^3+x^2}*\frac{2x+1}{x-3}=\frac{8(x-3)}{x^2(2x+1)}*\frac{2x+1}{x-3}=\frac{8}{x^2}*\frac{1}{1}=\frac{8}{x^2}[/tex]