ROWNANIA WYMIERNE :)Prosze o szybkie i dokladne rozwiazanie zadania 4.36 pp. a), b), c)trzeba podsta.... Question from @Kuba26841397522

Kuba26841397522 @Kuba26841397522

October 2018 1 17 Report

ROWNANIA WYMIERNE :)

Prosze o szybkie i dokladne rozwiazanie zadania 4.36 pp. a), b), c)

trzeba podstawic zmienną " t "

w 1 zał. zadania

w 2 zał. odpowiedzi

Roma

$\frac{3x^2 - 9x}{2} - \frac{12}{x^2-3x} = 3$

Zał.

$x^2-3x \neq 0 \\\ x \cdot (x - 3) \neq 0 \\\ x \neq 0 \wedge \ x - 3 \neq 0 \\\\ x \neq 0 \\\\ x - 3 \neq 0 \\\ x \neq 3 \\\\ D = R \setminus \{0; \ 3 \}$

$\frac{3x^2 - 9x}{2} - \frac{12}{x^2-3x} = 3 \\\\ \frac{3 \cdot (x^2 - 3x)}{2} - \frac{12}{x^2-3x} = 3 \\\\ t = x^2 - 3x \ \wedge \ x^2 - 3x \neq 0 \ => t \neq 0 \\\\ \frac{3t}{2} - \frac{12}{t} = 3 \\\\ \frac{3t}{2} - \frac{12}{t} - 3 = 0 \\\\ \frac{3t^2}{2t} - \frac{24}{2t} - \frac{6t}{2t} = 0 \\\\ \frac{3t^2 - 24 - 6t}{2t} = 0 \ / \cdot 2t \\\\ 3t^2 - 6t - 24 = 0 \ /:3 \\\\ t^2 - 2t - 8 = 0$

$\Delta = (-2)^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36; \ \sqrt{\Delta} = \sqrt{36} = 6 \\\\ t_1 = \frac{2 - 6}{2 \cdot 1} = \frac{-4}{2} = - 2 \\\\ t_2 = \frac{2 + 6}{2 \cdot 1} = \frac{8}{2} = 4 \\\\ t = x^2 - 3x \ \wedge \ t = -2 \\\ x^2 - 3x = - 2 \\\ x^2 - 3x + 2 = 0 \\\ \Delta = (-3)^2 - 4 \cdot 1 \cdot 2 =9 -8 = 1; \ \sqrt{\Delta} = \sqrt{1} = 1 \\\\ x_1 = \frac{3 - 1}{2 \cdot 1} = \frac{2}{2} = 1 \in D \\\\ x_2 = \frac{3 + 1}{2 \cdot 1} = \frac{4}{2} = 2 \in D$

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

Stąd:

$x \in \{-1; \ 1; \ 2; \ 4 \}$

$x^2 - 4x - \frac{15}{x^2 - 4x} = 2$

Zał.

$x^2 - 4x \neq 0 \\\ x \cdot (x - 4) \neq 0 \\\ x \neq 0 \ \wedge \ x - 4 \neq 0 \\\\ x \neq 0 \\\\ x - 4 \neq 0 \\\ x \neq 4 \\\\ D = R \setminus \{0; \ 4 \}$

$x^2 - 4x - \frac{15}{x^2 - 4x} = 2 \\\\ t = x^2 - 4x \ \wedge \ x^2 - 4x \neq 0 \ => \ t \neq 0 \\\\ t - \frac{15}{t} = 2 \\\\ t - \frac{15}{t} - 2 = 0 \\\\ \frac{t^2}{t} - \frac{15}{t} - \frac{2t}{t} = 0 \\\\ \frac{t^2 - 15 - 2t}{t} = 0 \ / \cdot t \\\\ t^2 - 2t - 15 = 0 \\\\ \Delta = (- 2)^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64; \ \sqrt{\Delta} = \sqrt{64} = 8 \\\\ t_1 = \frac{2 - 8}{2 \cdot 1} = \frac{- 6}{2} = -3 \\\\ t_2 = \frac{2 + 8}{2 \cdot 1} = \frac{10}{2} = 5$

$t = x^2 - 4x \ \wedge \ t = -3 \\\ x^2 - 4x = -3 \\\ x^2 - 4x +3 = 0 \\\ \Delta = (-4)^2 - 4 \cdot 1 \cdot 3 = 16 -12 = 4; \ \sqrt{\Delta} = \sqrt{4} = 2 \\\\ x_1 = \frac{4 - 2}{2 \cdot 1} = \frac{2}{2} = 1 \in D \\\\ x_2 = \frac{4 + 2}{2 \cdot 1} = \frac{6}{2} = 3 \in D$

$t = x^2 - 4x \ \wedge \ t = 5 \\\ x^2 - 4x = 5 \\\ x^2 - 4x - 5 = 0 \\\ \Delta = (-4)^2 - 4 \cdot 1 \cdot (-5) = 16 +20= 36; \ \sqrt{\Delta} = \sqrt{36} = 6 \\\\ x_1 = \frac{4 - 6}{2 \cdot 1} = \frac{-2}{2} = -1 \in D \\\\ x_2 = \frac{4 + 6}{2 \cdot 1} = \frac{10}{2} = 5 \in D$

Stąd:

$x \in \{-1; \ 1; \ 3; \ 5 \}$

$\frac{x^2-6x+3}{4} - \frac{5}{x^2-6x+3} = 2$

Zał.

$x^2-6x+3 \neq 0 \\\ \Delta = (-6)^2 - 4 \cdot 1 \cdot 3 = 36 - 12= 24; \ \sqrt{\Delta} = \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6}\\\\ x_1 = \frac{6 - 2\sqrt{6}}{2 \cdot 1} = \frac{2 \cdot (3- \sqrt{6})}{2} =3-\sqrt{6} \approx 0,55 \\\\ x_2 = \frac{6 + 2\sqrt{6}}{2 \cdot 1} = \frac{2 \cdot (3+ \sqrt{6})}{2} =3+\sqrt{6} \approx 5,45 \\\\ D = R \setminus \{3-\sqrt{6}; \ 3+\sqrt{6} \}$

$\frac{x^2-6x+3}{4} - \frac{5}{x^2-6x+3} = 2 \\\\ t = x^2-6x+3 \ \wedge \ x^2-6x+3 \neq 0 \ => \ t \neq 0 \\\\ \frac{t}{4} - \frac{5}{t} = 2 \\\\ \frac{t}{4} - \frac{5}{t} - 2 = 0 \\\\ \frac{t^2}{4t} - \frac{20}{4t} - \frac{8t}{4t} = 0 \\\\ \frac{t^2 - 20 - 8t}{4t} = 0 \ / \cdot 4t \\\\ t^2 - 8t - 20 = 0 \\\ \Delta = (-8)^2 - 4 \cdot 1 \cdot (-20) = 64 + 80 = 144; \ \sqrt{\Delta} = \sqrt{144} = 12 \\\\ t_1 = \frac{8-12}{2 \cdot 1} = \frac{-4}{2} = -2 \\\\ t_2 = \frac{8+12}{2 \cdot 1} = \frac{20}{2} = 10$

$t = x^2 - 6x+3 \ \wedge \ t = -2 \\\ x^2 - 6x +3= -2 \\\ x^2 - 6x +3+2=0 \\\ x^2 - 6x + 5 = 0 \\\ \Delta = (-6)^2 - 4 \cdot 1 \cdot 5 = 36 -20= 16; \ \sqrt{\Delta} = \sqrt{16} = 4 \\\\ x_1 = \frac{6 - 4}{2 \cdot 1} = \frac{2}{2} = 1 \in D \\\\ x_2 = \frac{6 + 4}{2 \cdot 1} = \frac{10}{2} = 5 \in D$

$t = x^2 - 6x+3 \ \wedge \ t = 10 \\\ x^2 - 6x +3= 10 \\\ x^2 - 6x +3-10=0 \\\ x^2 - 6x -7 = 0 \\\ \Delta = (-6)^2 - 4 \cdot 1 \cdot (-7) = 36 +28= 64; \ \sqrt{\Delta} = \sqrt{64} = 8 \\\\ x_1 = \frac{6 - 8}{2 \cdot 1} = \frac{-2}{2} = -1 \in D \\\\ x_2 = \frac{6 + 8}{2 \cdot 1} = \frac{14}{2} = 7 \in D$