Wykonaj działania...- Zadanie 9.15: Prosto do matury 2. Zakres podstawowy i rozszerzony

Rozwiązanie

$a) \frac{\frac{x + 1}{2 x - 4} + \frac{x - 5}{6 x - 12}}{\frac{x + 2}{3 x - 6} - \frac{x + 1}{4 x - 8}}$

Założenia:

$2 x - 4 \neq = 0 i 6 x - 12 \neq = 0 i 3 x - 6 \neq = 0 i 4 x - 8 \neq = 0 i \frac{x + 2}{3 x - 6} - \frac{x + 1}{4 x - 8} \neq = 0$

$2 (x - 2) \neq = 0 i 6 (x - 2) \neq = 0 i 3 (x - 2) \neq = 0 i 4 (x - 2) \neq = 0 i \frac{x + 2}{3 ( x - 2 )} - \frac{x + 1}{4 ( x - 2 )} \neq = 0$

$x - 2 \neq = 0 i \frac{4 \cdot ( x + 2 )}{12 ( x - 2 )} - \frac{3 ( x + 1 )}{12 ( x - 2 )} \neq = 0$

$x \neq = 2 i \frac{4 ( x + 2 ) - 3 ( x + 1 )}{12 ( x - 2 )} \neq = 0$

$x \neq = 2 i 4 x + 8 - 3 x - 3 \neq = 0$

$x \neq = 2 i x + 5 \neq = 0$

$x \neq = 2 i x \neq = - 5$

$x \in R - {- 5, 2}$

zatem:

$\frac{\frac{x + 1}{2 x - 4} + \frac{x - 5}{6 x - 12}}{\frac{x + 2}{3 x - 6} - \frac{x + 1}{4 x - 8}} = \frac{\frac{x + 1}{2 ( x - 2 )} + \frac{x - 5}{6 ( x - 2 )}}{\frac{x + 2}{3 ( x - 2 )} - \frac{x + 1}{4 ( x - 2 )}} = \frac{\frac{3 ( x + 1 ) + x - 5}{6 ( x - 2 )}}{\frac{4 ( x + 2 ) - 3 ( x + 1 )}{12 ( x - 2 )}} = \frac{\frac{3 x + 3 + x - 5}{6 x - 12}}{\frac{4 x + 8 - 3 x - 3}{12 x - 24}} = \frac{\frac{4 x - 2}{6 x - 12}}{\frac{x + 5}{12 x - 24}} = \frac{4 x - 2}{6 x - 12} \cdot \frac{12 x - 24}{x + 5} = \frac{2 ( 2 x - 1 )}{6 ( x - 2 )} \cdot \frac{12 \cdot ( 2 x - 2 )}{x + 5} = \frac{2 \cdot ( 2 x - 1 ) \cdot 2}{x + 5} = \frac{4 \cdot ( 2 x - 1 )}{x + 5} = \frac{8 x - 4}{x + 5}$

$b) \frac{\frac{x ^{2} + 10 x + 21}{x ^{2} + 8 x + 7} - 1}{\frac{x ^{2} - 4 x - 21}{x ^{2} - 6 x - 7} - 1}$

Założenia:

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i \frac{x ^{2} - 4 x - 21}{x ^{2} - 6 x - 7} - 1 \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i \frac{x ^{2} - 4 x - 21 - x ^{2} + 6 x + 7}{x ^{2} - 6 x - 7} \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i \frac{2 x - 14}{x ^{2} - 6 x - 7} \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i 2 x - 14 \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i 2 (x - 7) \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i x - 7 \neq = 0$

$x^{2} + 8 x + 7 \neq = 0 i x^{2} - 6 x - 7 \neq = 0 i x \neq = 7$

Rozłóżmy trójmian x²+8x+7 na czynniki:

$Δ = 8^{2} - 4 \cdot 1 \cdot 7 = 64 - 28 = 36, Δ = 6$

$x_{1} = \frac{- 8 - 6}{2} = \frac{- 14}{2} = - 7$

$x_{2} = \frac{- 8 + 6}{2} = \frac{- 2}{2} = - 1$

$x^{2} + 8 x + 7 = (x + 7) (x + 1)$

Rozłóżmy trójmian x²-6x-7 na czynniki:

$Δ = (- 6)^{2} - 4 \cdot 1 \cdot (- 7) = 36 + 28 = 64, Δ = 8$

$x_{1} = \frac{6 - 8}{2} = \frac{- 2}{2} = - 1$

$x_{2} = \frac{6 + 8}{2} = \frac{14}{2} = 7$

$x^{2} - 6 x - 7 = (x - 7) (x + 1)$

Wracamy do założeń:

$(x + 7) (x + 1) \neq = 0 i (x - 7) (x + 1) \neq = 0 i x \neq = 7$

$x + 7 \neq = 0 i x + 1 \neq = 0 i x - 7 \neq = 0 i x + 1 \neq = 0 i x \neq = 7$

$x \neq = - 7 i x \neq = - 1 i x \neq = 7$

zatem:

$\frac{\frac{x ^{2} + 10 x + 21}{x ^{2} + 8 x + 7} - 1}{\frac{x ^{2} - 4 x - 21}{x ^{2} - 6 x - 7} - 1} = \frac{\frac{x ^{2} + 10 x + 21 - x ^{2} - 8 x - 7}{x ^{2} + 8 x + 7}}{\frac{x ^{2} - 4 x - 21 - x ^{2} + 6 x + 7}{x ^{2} - 6 x - 7}} = \frac{\frac{2 x + 14}{x ^{2} + 8 x + 7}}{\frac{2 x - 14}{x ^{2} - 6 x - 7}} = \frac{2 x + 14}{x ^{2} + 8 x + 7} \cdot \frac{x ^{2} - 6 x - 7}{2 x - 14} = \frac{2 ( x + 7 ) \cdot ( x - 7 ) ( x + 1 )}{( x + 7 ) ( x + 1 ) \cdot 2 \cdot ( x - 7 )} = 1$

$c) \frac{x}{x + \frac{x}{x} + 1}$

Założenia:

$x + 1 \neq = 0 i x + \frac{x}{x + 1} \neq = 0$

$x \neq = - 1 i \frac{x ( x + 1 ) + x}{x + 1} \neq = 0$

$x \neq = - 1 i x^{2} + x + x \neq = 0$

$x \neq = - 1 i x^{2} + 2 x \neq = 0$

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

$x \neq = - 1 i x \neq = 0 i x + 2 \neq = 0$

$x \neq = - 1 i x \neq = 0 i x \neq = - 2$

zatem:

$\frac{x}{x + \frac{x}{x + 1}} = \frac{x}{\frac{x ( x + 1 ) + x}{x + 1}} = \frac{x}{\frac{x ^{2} + x + x}{x + 1}} = \frac{x}{\frac{x ^{2} + 2 x}{x + 1}} = x \cdot \frac{x + 1}{x ^{2} + 2 x} = \frac{x ^{2} + x}{x ^{2} + 2 x} = \frac{x ( x + 1 )}{x ( x + 2 )} = \frac{x + 1}{x + 2}$

$d) \frac{1}{1 - \frac{1}{1 - \frac{1}{x - 1}}}$

Założenia:

$x - 1 \neq = 0 i 1 - \frac{1}{x - 1} \neq = 0 i 1 - \frac{1}{1 - \frac{1}{x - 1}} \neq = 0$

$x \neq = 1 i \frac{x - 1 - 1}{x - 1} \neq = 0 i 1 - \frac{1}{x - 1 - 1} / (x - 1) \neq = 0$

$x \neq = 1 i x - 2 \neq = 0 i 1 - \frac{1}{\frac{x - 2}{x - 1}} \neq = 0$

$x \neq = 1 i x \neq = 2 i 1 - \frac{x - 1}{x - 2} \neq = 0$

$x \neq = 1 i x \neq = 2 i \frac{x - 2 - ( x - 1 )}{x - 2} \neq = 0$

$x \neq = 1 i x \neq = 2 i \frac{- 1}{x - 2} \neq = 0$

$x \neq = 1 i x \neq = 2$

zatem:

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