Question

Pls szybkoo pomoze ktos mam to na jutro rano zrobic

Answer 1

Odpowiedź:

Rozwiązanie: wystarczy pomnożyć przez potęgę mianownika. no i dziedzina oczywiście

a) D = R\{6}

(x+2)(x-6)=0

x = -2 v x ≠ 6

x = -2

b) D = R\{-3}

2(x-5)(x-6)(x+3)=0

x = 5 v x = 6 v x ≠ -3

x = 5 v x = 6

c) D = R\{[tex]\frac{3}{2}[/tex], 9}

3(2x-3)(9-x)(3-2x)=0

   -3(9-x)(2x-3)^2=0

x ≠ 9 v x ≠ [tex]\frac{3}{2}[/tex]

x = ∅

Answer 2

Verified answer

Zadanie 8.

a)

[tex]\frac{x+2}{x-6} = 0 \ \ \ |\cdot(x-6)\\\\Zal:\\x-6 \neq 0 \ \ \rightarrow \ \ x\neq 6\\D =R\setminus\{6\}\\\\\\x+2 = 0\\\\\boxed{x = -2}[/tex]

b)

[tex]\frac{(2x-10)(x-6)}{x+3} = 0 \ \ \ |\cdot(x+3)\\\\Zal:\\x+3 \neq 0 \ \ \rightarrow \ \ x \neq -3\\D = R \setminus\{-3\}\\\\\\(2x-10)(x-6) = 0\\\\2(x-5)(x-6) = 0 \ \ \ /:2\\\\(x-5)(x-6) = 0\\\\x-5 = 0 \ \vee \ x-6 = 0\\\\x = 5 \ \vee \ x = 6\\\\\boxed{x\in\{5,6\}}[/tex]

c)

[tex]\frac{2x-3}{(9-x)(9-6x)} = 0\\\\Zal:\\9-x \neq 0 \ \ \rightarrow \ \ x\neq 9\\i\\9-6x\neq 0 \ \ \rightarrow \ \ x \neq \frac{3}{2}\\D = R\setminus\{\frac{3}{2},9\}\\\\\\2x-3 = 0\\\\2x = 3 \ \ \ /:2\\\\x = \frac{3}{2} \ \ \notin D\\\\x = \emptyset[/tex]