Odpowiedź :
Odpowiedź:
Zadanie 2.
Odpowiedź : x=0 v x=-3 v x= -4
Szczegółowe wyjaśnienie:
x(x+3)(2x+8)=0
x(x+3)2(x+4)=0
x= 0 v x+3=0 v x+4=0
x=0 v x=-3 v x= -4
2.
[tex]x(x+3)(2x+8) = 0\\\\2x(x+3)(x+4) = 0 \ \ /:2\\\\x(x+3)(x+4) = 0\\\\x = 0 \ \vee \ x + 3 = 0 \ \vee \ x +4 = 0\\\\x = 0 \ \vee \ x = -3 \ \vee \ x = -4\\\\x \in \{-4, -3, 0}\}[/tex]
3.
[tex]a) \ \ x \neq 0\\\\\frac{6}{x} + \frac{5}{2x} = \frac{2\cdot6}{2x}+\frac{5}{2x} = \frac{12+5}{2x} = \frac{17}{2x}\\\\b) \ \ x\neq -3 \ i \ x \neq 0\\\\\frac{6x}{x+3}\cdot\frac{x+3}{x^{2}} = \frac{6x}{x^{2}} = \frac{6}{x}[/tex]
4.
[tex]a) \ log_{6}4 + log_{6}9 = log_{6}(4\cdot9) = log_{6}36 = log_{6}6^{2} = 2\\\\b) \ log_{3}18-log_{3}2 = log_{3}\frac{18}{2} = log_{3}9 = log_{3}3^{2} = 2[/tex]
5.
[tex]2,4,6,8,...\\a_1 = 2\\r = a_2 - a_1 = 4-2 = 2\\n = 6\\S_{6} = ?\\\\a_{n} = a_1+(n-1)\cdot r\\\\a_{6} = 2+5\cdot 2 = 2+10 = 12\\\\S_{n} = \frac{a_1+a_{n}}{2}\cdot n\\\\S_{6} = \frac{2+12}{2}\cdot 6 = 14\cdot3 = 42[/tex]
6.
[tex]1,3,9,27,...\\a_1 = 1\\a_2 = 3\\q = ?\\a_5, \ a_6 = ?\\\\q = \frac{a_2}{a_1} = \frac{3}{1} = 3\\\\a_{n} = a_1\cdot q^{n-1}\\\\a_5 = a_1\cdot q^{4} = 1\cdot3^{4} = 81\\\\a_6 = a_1\cdot q^{5} = 1\cdot3^{5} = 243[/tex]