Question

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

Answer 1

Odpowiedź:

a)

[tex]\frac{2x}{x+1}+\frac{x-1}{x}=\frac{2x^2}{x(x+1)}+\frac{(x-1)(x+1)}{x(x+1)}=\frac{2x^2+x^2-1}{x(x+1)}=\frac{3x^2-1}{x^2+x}\\\\x(x+1)\neq0\\x\neq0 \ \wedge \ x\neq -1\\ D = \mathbb{R} \setminus\lbrace-1,0\rbrace[/tex]

b)

[tex]\frac{x^2-16}{x^2+x}-\frac{1}{x^2-1}=\frac{x^2-16}{x(x+1)}-\frac{1}{(x-1)(x+1)}=\frac{(x^2-16)(x-1)}{x(x-1)(x+1)}-\frac{x}{x(x-1)(x+1)}=\frac{x^3-x^2-16x+16-x}{x(x+1)(x-1)}=\frac{x^3-x^2-17x+16}{x^3-x}\\\\x(x-1)(x+1)\neq0\\x\neq0 \ \wedge \ x\neq-1 \ \wedge \ x\neq1\\\\D=\mathbb{R}\setminus\lbrace-1,0,1\rbrace[/tex]

c)

[tex]\frac{3x^2}{x+4}*\frac{x^2-16}{6x^2-3x}=\frac{3x^2}{x+4}*\frac{(x+4)(x-4)}{3x(2x-1)}=\frac{x}{1}*\frac{x-4}{2x-1}=\frac{x(x-4)}{2x-1}=\frac{x^2-4x}{2x-1}\\\\x+4\neq0 \ \wedge \ 6x^2-3x\neq0\\x\neq -4 \ \ \ \ \ \ \ \ 3x(2x-1)\neq0\\.. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\neq0 \ \wedge \ x\neq\frac{1}{2}\\\\D=\mathbb{R} \setminus \lbrace-4,0,\frac{1}{2}\rbrace[/tex]

d)

[tex]\frac{5x^3-15x^2}{3x^2-12}:\frac{4x-12}{x-2}=\frac{5x^2(x-3)}{3(x^2-4)}:\frac{4x-12}{x-2}=\frac{5x^2(x-3)}{3(x+2)(x-2)}*\frac{x-2}{4(x-3)}=\frac{5x^2}{3(x+2)}*\frac{1}{4}=\frac{5x^2}{12(x+2)}=\frac{5x^2}{12x+24}\\\\3x^2-12\neq0 \ \wedge \ x-2\neq0 \ \wedge \ 4x-12\neq0\\3x^2\neq12 \ \ \ \ \ \ \ \ \ \ \ x\neq2 \ \ \ \ \ \ \ \ \ \ 4x\neq12 \\ x^2\neq4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\neq 3 \\x\neq2 \ \wedge \ x\neq-2\\\\D=\mathbb{R} \setminus \lbrace-2,2,3\rbrace[/tex]