zrobilem te zadania ale gdyby ktos mogl je zrobic to mogłbym sprawdzic czy mam dobrze ajbardziej mi .... Question from @patrykbons

November 2018 1 31 Report

zrobilem te zadania ale gdyby ktos mogl je zrobic to mogłbym sprawdzic czy mam dobrze ajbardziej mi zalezy na 51, 52 53 54 55 bo tych nie jestem pewny ale gdyby wszystkie to byloby ok

Roma

Zad. 47 - 52 w załączniku

Zad. 53

$3x^2 + 14x - 5 \geq 0 \\\ 3x^2 + 14x - 5 = 0 \\\ \Delta = 14^2 - 4 \cdot 3 \cdot (-5) = 196 + 60 = 256 \\\\ x_1 = \frac{-14- \sqrt{256}}{2 \cdot 3} = \frac{-14-16}{6}=\frac{-30}{6} = -5 \\\\ x_2 = \frac{-14+\sqrt{256}}{2 \cdot 3} = \frac{-14+16}{6}=\frac{2}{6} =\frac{1}{3} \\\\ x \in (-\infty; \ -5 \rangle \cup \langle \frac{1}{3}; \ +\infty)$

$-2x^2 - 7x + 4 \leq 0 \\\ -2x^2 - 7x + 4 = 0 \\\ \Delta = (-7)^2 - 4 \cdot (-2) \cdot 4= 49 + 32 = 81 \\\\ x_1 = \frac{7- \sqrt{81}}{2 \cdot (-2)} = \frac{7-9}{-4}=\frac{-2}{-4} = \frac{1}{2} \\\\ x_2 = \frac{7+ \sqrt{81}}{2 \cdot (-2)} = \frac{7+9}{-4}=\frac{16}{-4} =-4 \\\\ x \in (-\infty; \ -4 \rangle \cup \langle \frac{1}{2}; \ +\infty)$

$3x^2 + 3x + 1 < 0 \\\ 3x^2 + 3x + 1 = 0 \\\ \Delta = 3^2 - 4 \cdot 3 \cdot 1 = 9 -12 = - 3 < 0 \\\\ x \in \emptyset$

$x^2 + x - 1 > 0 \\\ x^2 + x - 1 = 0 \\\ \Delta = 1^2 - 4 \cdot 1 \cdot (-1) = 1 + 4= 5 \\\\ x_1 = \frac{-1- \sqrt{5}}{2 \cdot 1} = \frac{-1- \sqrt{5}}{2} \\\\ x_2 = \frac{-1+\sqrt{5}}{2 \cdot 1} = \frac{-1+ \sqrt{5}}{2} \\\\ x \in (-\infty; \ \frac{-1- \sqrt{5}}{2}) \cup (\frac{-1+ \sqrt{5}}{2}; \ +\infty)$

$3x-x^2 \leq 3 - 3x^2 \\\ 3x-x^2-3 + 3x^2 \leq 0 \\\ 2x^2+3x-3 \leq 0 \\\ 2x^2+3x-3 =0 \\\ \Delta = 3^2 - 4 \cdot 2 \cdot (-3) = 9 + 24 = 33 \\\\ x_1 = \frac{-3- \sqrt{33}}{2 \cdot 2} = \frac{-3- \sqrt{33}}{4} \\\\ x_2 = \frac{-3+\sqrt{33}}{2 \cdot 2} = \frac{-3+ \sqrt{33}}{4}\\\\ x \in ( \frac{-3- \sqrt{33}}{4}; \ \frac{-3+ \sqrt{33}}{4})$

$-x(2 -x) > 1 - x^2 \\\ -2x+ x^2 > 1 - x^2 \\\ -2x+ x^2 -1+x^2 >0 \\\ 2x^2-2x - 1 > 0 \\\ \Delta = (-2)^2 - 4 \cdot 2 \cdot (-1) = 4 + 8 = 12 \\\\ x_1 = \frac{2- \sqrt{12}}{2 \cdot 2} = \frac{2- \sqrt{4\cdot 3}}{4}= \frac{2- 2\sqrt{3}}{4} = \frac{2 \cdot (1- \sqrt{3})}{4} = \frac{1- \sqrt{3}}{2} \\\\ x_2 = \frac{2+ \sqrt{12}}{2 \cdot 2} = \frac{2+ \sqrt{4\cdot 3}}{4}= \frac{2+2\sqrt{3}}{4} = \frac{2 \cdot (1+ \sqrt{3})}{4} = \frac{1+ \sqrt{3}}{2}$

$x \in (-\infty; \ \frac{1- \sqrt{3}}{2}) \cup (\frac{1+ \sqrt{3}}{2}; \ +\infty)$

Zad. 54

$(2x + 1)^2 + (x- 3)^2 < 10 \\\ 4x^2 + 4x + 1 + x^2 - 6x + 9 - 10 < 0 \\\ 5x^2- 2x < 0 \\\ 5x^2 - 2x = 0 \\\ x \cdot (5x - 2) = 0 \\\ x = 0 \ lub \ 5x - 2 = 0 \\\ x = 0 \ lub \ 5x = 2 \ /:5 \\\ x = 0 \ lub \ x = \frac{2}{5} \\\ x \in (0; \ \frac{2}{5})$

$3x - (1 - x)^2 \geq (x - 2)(x+ 2) \\\ 3x -(1 - 2x + x^2) \geq x^2 - 4 \\\ 3x - 1 + 2x -x^2 - x^2 + 4 \geq 0 \\\ -2x^2 + 5x + 3 \geq 0 \\\ \Delta = 5^2 - 4 \cdot (-2) \cdot 3= 25+24 = 49 \\\\ x_1= \frac{-5 - \sqrt{49}}{2 \cdot (-2)} =\frac{-5-7}{-4}=\frac{-12}{-4} = 3 \\\\ x_2= \frac{-5 + \sqrt{49}}{2 \cdot (-2)} =\frac{-5+7}{-4}=\frac{2}{-4} = -\frac{1}{2} \\\\ x \in \langle - \frac{1}{2}, \ 3 \rangle$

$(\frac{1}{3}x+ \frac{1}{3})^2 > \frac{2}{3}x- \frac{8}{3} \\\\ \frac{1}{9}x^2 +\frac{2}{9}x + \frac{1}{9} >\frac{4}{9}x-\frac{24}{9} \\\\ \frac{1}{9}x^2 +\frac{2}{9}x + \frac{1}{9} -\frac{4}{9}x+\frac{24}{9} > 0 \\\\ \frac{1}{9}x^2 -\frac{2}{9}x + \frac{25}{9} > 0 \ / \cdot 9 \\\\ x^2 - 2x + 25 > 0 \\\\ x^2 - 2x + 25 = 0 \\\\ \Delta = (-4)^2 - 4 \cdot 1 \cdot 25 = 16 - 100 = - 84 < 0 \\\\ x \in R$

$2 - x^2 < (2 - x)^2 \\\ 2 - x^2 < 4 - 4x + x^2 \\\ 2 - x^2 - 4+ 4x - x^2 < 0 \\\ - 2x^2 + 4x - 2 < 0 \:(-2) \\\ x^2 - 2x + 1 > 0 \\\ (x - 1)^2 > 0 \\\ (x - 1)^2 = 0 \ /^{\sqrt{}} \\\ x - 1 = 0 \\\ x = 1 \\\ x \in (-\infty; \ 1) \cup (1; \ + \infty)$

Zad. 55

$x^2 - 4x +1 \leq 0 \\\ x^2 - 4x +1 =0 \\\ \Delta = (-4)^2 - 4 \cdot 1 \cdot 1= 16-4 = 12 \\\\ x_1= \frac{4 - \sqrt{12}}{2 \cdot 1} =\frac{4 - \sqrt{4 \cdot 3}}{2} =\frac{4 - 2\sqrt{3}}{2} =\frac{2 \cdot (2 - \sqrt{3})}{2} =2 - \sqrt{3} \approx 0,27\\\\ x_2=\frac{4 + \sqrt{12}}{2 \cdot 1} =\frac{4+ \sqrt{4 \cdot 3}}{2} =\frac{4 + 2\sqrt{3}}{2} =\frac{2 \cdot (2 + \sqrt{3})}{2} =2 + \sqrt{3} \approx 3,73\\\\ x \in \langle 2 - \sqrt{3}; \ 2 + \sqrt{3} \rangle$

Zatem:

$A = \langle 2 - \sqrt{3}; \ 2 + \sqrt{3} \rangle$

$3x - x^2 \geq 0 \\\ -x^2 + 3x \geq 0 \\\ -x^2 + 3x = 0 \\\ -x \cdot (x - 3) = 0 \\\ -x = 0 \ / \cdot (-1) \ lub \ x - 3 = 0 \\\ x = 0 \ lub \ x= 3 \\\ x \in \langle 0; \ 3 \rangle$

Zatem:

$B = \langle 0; \ 3 \rangle$

Stąd:

$A = \langle 2 - \sqrt{3}; \ 2 + \sqrt{3} \rangle , \ B = \langle 0; \ 3 \rangle \\\\ A \cap B = [\langle 2 - \sqrt{3}; \ 2 + \sqrt{3} \rangle \cap \langle 0; \ 3 \rangle] = \langle 2 - \sqrt{3}; \ 3 \rangle \\\\ A \setminus B = [\langle 2 - \sqrt{3}; \ 2 + \sqrt{3} \rangle \setminus \langle 0; \ 3 \rangle] = (3; \ 2 + \sqrt{3} \rangle$

$x^2 + 3x + 1 > 0 \\\ x^2 + 3x + 1 = 0 \\\ \Delta = 3^2 - 4 \cdot 1 \cdot 1 = 9-4 = 5 \\\\ x_1 = \frac{-3- \sqrt{5}}{2 \cdot 1} = \frac{-3- \sqrt{5}}{2} \approx -2,62\\\\ x_2 =\frac{-3+\sqrt{5}}{2 \cdot 1} = \frac{-3+\sqrt{5}}{2} \approx - 0,38 \\\\ x \in (-\infty; \ \frac{-3- \sqrt{5}}{2}) \cup ( \frac{-3+\sqrt{5}}{2}; \ +\infty)$

Zatem:

$A = (-\infty; \ \frac{-3- \sqrt{5}}{2}) \cup ( \frac{-3+\sqrt{5}}{2}; \ +\infty)$

$x^2 - 5x + 4 < 0 \\\ x^2 - 5x + 4 = 0 \\\ \Delta = (-5)^2 - 4 \cdot 1 \cdot 4= 25-16= 9 \\\\ x_1= \frac{5 - \sqrt{9}}{2 \cdot 1} =\frac{5-3}{2}=\frac{2}{2} = 1 \\\\ x_2= \frac{5 + \sqrt{9}}{2 \cdot 1} =\frac{5+3}{2}=\frac{8}{2} = 4 \\\\ x \in (1; \ 4)$

Zatem:

$B =(1; \ 4)$

Stąd:

$A = (-\infty; \ \frac{-3- \sqrt{5}}{2}) \cup ( \frac{-3+\sqrt{5}}{2}; \ +\infty), \ B = (1; \ 4) \\\\ A \cap B = \{[(-\infty; \ \frac{-3- \sqrt{5}}{2}) \cup ( \frac{-3+\sqrt{5}}{2}; \ +\infty)] \cap (1; \ 4) \}= (1; \ 4) = B \\\\ A \setminus B = \{[(-\infty; \ \frac{-3- \sqrt{5}}{2})\cup ( \frac{-3+\sqrt{5}}{2}; \ +\infty)] \setminus (1; \ 4) \}= \{1; \ 4\}$