Rozwiąż równanie.- Zadanie 1.8: Prosto do matury 3. Zakres podstawowy i rozszerzony. Po gimnazjum

Rozwiązanie

$a)$

$\frac{3 x - 1}{5 - x} - \frac{x}{x - 2} = 0$

Zał:

$5 - x \neq = 0, x - 2 \neq = 0$

$5 \neq = x, x \neq = 2$

$\frac{3 x - 1}{5 - x} = \frac{x}{x - 2}$

$(3 x - 1) (x - 2) = x (5 - x)$

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

$4 x^{2} - 12 x + 2 = 0 ∣ : 2$

$2 x^{2} - 6 x + 1 = 0$

$Δ = (- 6)^{2} - 4 \cdot 2 \cdot 1 = 36 - 8 = 28$

$Δ = 28 = 27$

$x_{1} = \frac{6 - 2 7}{4} = \frac{3 - 7}{2}$

$x_{2} = \frac{6 + 2 7}{4} = \frac{3 + 7}{2}$

$b)$

$\frac{x ^{2} + 4 x - 14}{x ^{2} - 4 x + 6} = - 3$

Zał:

$x^{2} - 4 x + 6 \neq = 0$

$Δ = (- 4)^{2} - 4 \cdot 1 \cdot 6 = 16 - 36 < 0$

Zatem:

$x \in R$

$x^{2} + 4 x - 14 = - 3 (x^{2} - 4 x + 6)$

$x^{2} + 4 x - 14 = - 3 x^{2} + 12 x - 18$

$4 x^{2} - 8 x + 4 = 0 ∣ : 4$

$x^{2} - 2 x + 1 = 0$

$(x - 1)^{2} = 0$

$x - 1 = 0$

$x = 1$

$c)$

$2 + \frac{3 x + 7}{3 x - 5} = \frac{6 x + 6}{3 x - 1} + \frac{9 x ^{2} + 15 x - 30}{9 x ^{2} - 18 x + 5}$

Zał:

$3 x - 5 \neq = 0, 3 x - 1 \neq = 0, 9 x^{2} - 18 x + 5 \neq = 0$

$x \neq = \frac{5}{3}, x \neq = \frac{1}{3}, Δ = (- 18)^{2} - 4 \cdot 9 \cdot 5 = 324 - 180 = 144$

$x \neq = \frac{5}{3}, x \neq = \frac{1}{3}, x \neq = \frac{18 - 12}{18} = \frac{1}{3}, x \neq = \frac{18 + 12}{18} = \frac{30}{18} = \frac{5}{3}$

$2 + \frac{3 x + 7}{3 x - 5} = \frac{6 x + 6}{3 x - 1} + \frac{9 x ^{2} + 15 x - 30}{( 3 x - 5 ) ( 3 x - 1 )} ∣ \cdot (3 x - 1) (3 x - 5)$

$2 (9 x^{2} - 18 x + 5) + (3 x + 7) (3 x - 1) = (6 x + 6) (3 x - 5) + 9 x^{2} + 15 x - 30$

$18 x^{2} - 36 x + 10 + 9 x^{2} - 3 x + 21 x - 7 = 18 x^{2} - 30 x + 18 x - 30 + 9 x^{2} + 15 x - 30$

$- 18 x + 3 = 3 x - 60$

$- 21 x = - 63 ∣ : (- 21)$

$x = 3$

$d)$

$\frac{3 x + 1}{x ^{2} - 4 x + 3} - \frac{12 x - 10}{x ^{3} - 2 x ^{2} - 5 x + 6} = 1 - \frac{x ^{2} - 3 x - 3}{x ^{2} + x - 2}$

Zał:

$1)$

$x^{2} - 4 x + 3 \neq = 0$

$Δ = (- 4)^{2} - 4 \cdot 1 \cdot 3 = 16 - 12 = 4$

$x \neq = \frac{4 - 2}{2} = \frac{2}{2} = 1$

$x \neq = \frac{4 + 2}{2} = \frac{6}{2} = 3$

$2)$

$x^{3} - 2 x^{2} - 5 x + 6 \neq = 0$

$x^{3} - x^{2} - 6 x - x^{2} + x + 6 \neq = 0$

$x (x^{2} - x - 6) - 1 (x^{2} - x - 6) \neq = 0$

$(x - 1) (x^{2} - x - 6) \neq = 0$

$(x - 1) (x^{2} - 3 x + 2 x - 6) \neq = 0$

$(x - 1) (x + 2) (x - 3) \neq = 0$

$x - 1 \neq = 0, x + 2 \neq = 0, x - 3 \neq = 0$

$x \neq = 1, x \neq = - 2, x \neq = 3$

$3)$

$x^{2} + x - 2 \neq = 0$

$Δ = 1^{2} - 4 \cdot 1 \cdot (- 2) = 1 + 8 = 9$

$x \neq = \frac{- 1 - 3}{2} = \frac{- 4}{2} = - 2$

$x \neq = \frac{- 1 + 3}{2} = \frac{2}{2} = 1$

Zatem:

$D = R - {- 2, 1, 3}$

$\frac{3 x + 1}{x ^{2} - 4 x + 3} - \frac{12 x + 10}{x ^{3} - 2 x ^{2} - 5 x + 6} = 1 - \frac{x ^{2} - 3 x - 3}{x ^{2} + x - 2}$

$\frac{3 x + 1}{( x - 1 ) ( x - 3 )} - \frac{12 x + 10}{( x - 1 ) ( x + 2 ) ( x - 3 )} = 1 - \frac{x ^{2} - 3 x - 3}{( x + 2 ) ( x - 1 )} ∣ \cdot (x - 1) (x + 2) (x - 3)$

$(3 x + 1) (x + 2) - (12 x + 10) = (x - 1) (x + 2) (x - 3) - (x^{2} - 3 x - 3) (x - 3)$

$3 x^{2} + 6 x + x + 2 - 12 x - 10 = x^{3} - 2 x^{2} - 5 x + 6 - (x^{3} - 3 x^{2} - 3 x^{2} + 9 x - 3 x + 9)$

$3 x^{2} - 5 x - 8 = x^{3} - 2 x^{2} - 5 x + 6 - x^{3} + 3 x^{2} + 3 x^{2} - 9 x + 3 x - 9$

$- 8 = x^{2} - 6 x - 3$

$x^{2} - 6 x + 5 = 0$

$Δ = (- 6)^{2} - 4 \cdot 1 \cdot 5 = 36 - 20 = 16$

$x_{1} = \frac{6 - 4}{2} = \frac{2}{2} = 1 \in / D$

$x_{2} = \frac{6 + 4}{2} = \frac{10}{2} = 5$

Magdalena Matusik

Nauczycielka matematyki

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