Odpowiedź:
zad. 1
[tex]a_{n} = - 2n + 5 \\ a_{1} = - 2 \times 1 + 5 = - 2 + 5 = 3 \\ a_{2} = - 2 \times 2 + 5 = - 4 + 5 = 1 \\ a_{3} = - 2 \times 3 + 5 = - 6 + 5 = - 1 \\ a_{4} = - 2 \times 4 + 5 = - 8 + 5 = - 3 \\ a_{3} - a_{2} = a_{2} - a_{1} \\ - 1 - 1 = 1 - 3 \\ - 2 = - 2[/tex]
jest to ciąg arytmetyczny o różnicy -2
jest to ciąg malejący
zad. 2
[tex] log_{0.5}(3 - 2x) = - 1 \: \: \: \: \: x < \frac{3}{2} \\ 3 - 2x = {0.5}^{ - 1} \\ 3 - 2x = {( \frac{1}{2}) }^{ - 1} \\ 3 - 2x = 2 \\ - 2x = 2 - 3 \\ - 2x = - 1 \\ x = \frac{1}{2} \: \: \: \: \: \: \: x < \frac{3}{2} \\ \\ x = \frac{1}{2} [/tex]
zad. 3
[tex] \frac{1}{x} + \frac{1}{y + 1} = \frac{1}{a} \\ y + 1≠0 \\ y≠ - 1 \\ \\ \frac{1}{y + 1} = \frac{1}{a} - \frac{1}{x} \: \: \: | \div 1 \\ y + 1 = \frac{1}{ \frac{1}{a} - \frac{1}{x} } \\ y + 1 = \frac{1}{ \frac{x - a}{ax} } \\ y + 1 = \frac{ax}{x - a} \\ y = \frac{ax}{x - a} - 1[/tex]
(nie jestem pewna czy dobrze jest)
zad. 4
[tex] {9}^{1 + \sqrt{27}} \div {27}^{1 + 2 \sqrt{3} } = {9}^{1 + 3 \sqrt{3} } \div {3}^{3 + 6 \sqrt{3} } = {3}^{2 + 6 \sqrt{3} } \div {3}^{3 + 6 \sqrt{3} } = {3}^{ - 1} = \frac{1}{3} [/tex]
[tex]3 {x}^{3} - {x}^{2} + ax + 4 = 0 \\ 3 \times ( \frac{1}{3} {)}^{3} - ( { \frac{1}{3}) }^{2} + \frac{1}{3} a + 4 = 0 \\ 3 \times \frac{1}{27} - \frac{1}{9} + \frac{1}{3} a + 4 = 0 \\ \frac{1}{9} - \frac{1}{9} + \frac{1}{3} a + 4 = 0 \\ \frac{1}{3} a + 4 = 0 \\ \frac{1}{3} a = - 4 \: \: \: \: \: | \times 3 \\ a = - 12[/tex]
[tex]3 {x}^{3} - {x}^{2} - 12x + 4 = 0 \\ {x}^{2} (3x - 1) - 4(3x - 1) = 0 \\ (3x - 1)( {x}^{2} - 4) = 0 \\ \\ 3x - 1 = 0 \\ x = \frac{1}{3} \\ \\ {x}^{2} - 4 = 0 \\ {x}^{2} = 4 \\ x = - 2 \: \: \: \: \: x = 2 \\ \\x = - 2 \: \: \: \: \: x = 2 \: \: \: \: \: \: \: x = \frac{1}{3} [/tex]
